MATH 120A COMPLEX VARIABLES NOTES: …bdriver/120A-F03/Lecture Notes/Fall03-Lecture...MATH 120A...

MATH 120A COMPLEX VARIABLES NOTES: REVISEDDecember 3, 2003

BRUCE K. DRIVER

Abstract. Math 120A Lecture notes, Fall 2003.

Date : File:complex.tex Last revised: December 3, 2003.

1

2 BRUCE K. DRIVER

1. (9/26/03)

1.1. Introduction. For our purposes the definition of complex variables is thecalculus of analytic functions, where a function F (x, y) = (u(x, y), v(x, y)) fromR2 to itself is analytic iff it satisfies the Cauchy Riemann equations:

ux = −vy and vx = uy.

Because this class of functions is so restrictive, the associated calculus has some verybeautiful and useful properties which will be explained in this class. The followingfact makes the subject useful in applications.

Fact 1.1. Many of the common elementary functions, like xn, ex, sinx, tanx, lnx,etc. have unique “extensions” to analytic functions. Moreover, the solutions tomany ordinary differential equations extend to analytic functions. So the study ofanalytic functions aids in understanding these class of real valued functions.

1.2. Book Sections 1-5.

Definition 1.2 (Complex Numbers). Let C = R2 equipped with multiplicationrule

(1.1) (a, b)(c, d) ≡ (ac− bd, bc+ ad)

and the usual rule for vector addition. As is standard we will write 0 = (0, 0) ,1 = (1, 0) and i = (0, 1) so that every element z of C may be written as z = x1+ yiwhich in the future will be written simply as z = x+ iy. If z = x+ iy, let Re z = xand Im z = y.

Writing z = a+ ib and w = c+ id, the multiplication rule in Eq. (1.1) becomes

(1.2) (a+ ib)(c+ id) ≡ (ac− bd) + i(bc+ ad)

and in particular 12 = 1 and i2 = −1.Proposition 1.3. The complex numbers C with the above multiplication rule sat-isfies the usual definitions of a field. For example wz = zw and z (w1 + w2) =zw1 + zw2, etc. Moreover if z 6= 0, z has a multiplicative inverse given by

(1.3) z−1 =a

a2 + b2− i

b

a2 + b2.

Probably the most painful thing to check directly is the associative law, namely

(1.4) u (vw) = (uv)w.

This can be checked later in polar form easier.

Proof. Suppose z = a + ib 6= 0, we wish to find w = c + id such that zw = 1and this happens by Eq. (1.2) iff

ac− bd = 1 and(1.5)

bc+ ad = 0.(1.6)

Now taking a(1.5) + b (1.6) implies¡a2 + b2

¢c = a and so c = a

a2+b2 and taking−b(1.5) + a (1.6) implies

¡a2 + b2

¢d = −b and hence c = − b

a2+b2 as claimed.

MATH 120A COMPLEX VARIABLES NOTES: REVISED December 3, 2003 3

Remark 1.4 (Not Done in Class). Here is a way to understand some of the basicproperties of C using our knowledge of linear algebra. LetMz denote multiplicationby z = a+ ib then if w = c+ id we have

Mzw =

µac− bdbc+ ad

¶=

µa −bb a

¶µcd

¶

so that Mz =

µa −bb a

¶= aI + bJ where J :=

µ0 −11 0

¶. With this notation

we have MzMw = Mzw and since I and J commute it follows that zw = wz.Moreover, since matrix multiplication is associative so is complex multiplication,i.e. Eq. (1.4) holds. Also notice thatMz is invertible iff detMz = a2+b2 = |z|2 6= 0in which case

M−1z =1

|z|2µ

a b−b a

¶=Mz/|z|2

as we have already seen above.

Notation 1.5. We will write 1/z for z−1 and w/z to mean z−1 · w.Notation 1.6 (Conjugation and Modulous). If z = a+ib with a, b ∈ R let z = a−iband

|z|2 ≡ zz = a2 + b2.

Notice that

(1.7) Re z =1

2(z + z) and Im z =

1

2i(z − z) .

Proposition 1.7. Complex conjugation and the modulus operators satisfy,

(1) z = z,(2) zw = zw and z + w = z + w.(3) |z| = |z|(4) |zw| = |z| |w| and in particular |zn| = |z|n for all n ∈ N.(5) |Re z| ≤ |z| and |Im z| ≤ |z|(6) |z + w| ≤ |z|+ |w| .(7) z = 0 iff |z| = 0.(8) If z 6= 0 then z−1 := z

|z|2 (also written as1z ) is the inverse of z.

(9)¯z−1

¯= |z|−1 and more generally |zn| = |z|n for all n ∈ Z.

Proof. 1. and 3. are geometrically obvious.2. Say z = a+ ib and w = c+ id, then zw is the same as zw with b replaced by

−b and d replaced by −d, and looking at Eq. (1.2) we see thatzw = (ac− bd)− i(bc+ ad) = zw.

4. |zw|2 = zwzw = zzww = |z|2 |w|2 as real numbers and hence |zw| = |z| |w| .5. Geometrically obvious or also follows from

|z| =q|Re z|2 + |Im z|2.

4 BRUCE K. DRIVER

6. This is the triangle inequality which may be understood geometrically or bythe computation

|z + w|2 = (z + w) (z + w) = |z|2 + |w|2 + wz + wz

= |z|2 + |w|2 + wz + wz

= |z|2 + |w|2 + 2Re (wz) ≤ |z|2 + |w|2 + 2 |z| |w|= (|z|+ |w|)2 .

7. Obvious.8. Follows from Eq. (1.3).

9.¯z−1

¯=¯z|z|2¯=¯1|z|2¯|z| = 1

|z| .


2. (9/30/03)

2.1. Left Overs. Go over Eq. (1.7) and properties 8. and 9. in Proposition 1.7.

Lemma 2.1. For complex number u, v,w, z ∈ C with v 6= 0 6= z, we have

1

u

1

v=1

uv, i.e. u−1v−1 = (uv)−1

u

v

w

z=

uw

vzand

u

v+

w

z=

uz + vw

vz.

Proof. For the first item, it suffices to check that

(uv)¡u−1v−1

¢= u−1uvv−1 = 1 · 1 = 1.

The rest follow usingu

v

w

z= uv−1wz−1 = uwv−1z−1 = uw (vz)−1 =

uw

vz.

u

v+

w

z=

z

z

u

v+

v

v

w

z=

zu

zv+

vw

vz

= (vz)−1(zu+ vw) =

uz + vw

vz.

2.2. Book Sections 36-37, p. 111-115. Here we suppose w (t) = c (t) + id (t)and define

w (t) = c (t) + id (t)

and Z β

α

w (t) dt :=

Z β

α

c (t) dt+ i

Z β

α

d (t) dt

Example 2.2. Z π/2

0

¡et + i sin t

¢dt = e

12π − 1 + i.

Theorem 2.3. If z (t) = a (t) + ib (t) and w (t) = c (t) + id(t) and λ = u+ iv ∈ Cthen

(1) ddt (w (t) + z (t)) = w (t) + z (t)

(2) ddt [w (t) z (t)] = wz + wz

(3)R βα[w (t) + λz (t)] dt =

R βαw (t) dt+ λ

R βαz (t) dt

(4)R βαw(t)dt = w(β)− w(α) In particular if w = 0 then w is constant.

(5) Z β

α

w(t)z(t)dt = −Z β

α

w(t)z(t)dt+ w (t) z (t) |βα.(6) ¯

¯Z β

α

w (t) dt

¯¯ ≤

Z β

α

|w (t)| dt.

6 BRUCE K. DRIVER

Proof. 1. and 4. are easy.2.

d

dt[wz] =

d

dt(ac− bd) + i

d

dt(bc+ ad)

= (ac− bd) + i(bc+ ad)

+ (ac− bd) + i(bc+ ad)

= wz + wz.

3. The only interesting thing to check is thatZ β

α

λz (t) dt = λ

Z β

α

z (t) dt.

Again we simply write out the real and imaginary parts:

Z β

α

λz (t) dt =

Z β

α

(u+ iv) (a (t) + ib (t)) dt

=

Z β

α

(ua(t)− vb(t) + i [ub(t) + va(t)]) dt

=

Z β

α

(ua(t)− vb(t)) dt+ i

Z β

α

[ub(t) + va(t)] dt

while Z β

α

λz (t) dt = (u+ iv)

Z β

α

[a (t) + ib (t)] dt

= (u+ iv)

ÃZ β

α

a (t) dt+ i

Z β

α

b (t) dt

!

=

Z β

α

(ua(t)− vb(t)) dt+ i

Z β

α

[ub(t) + va(t)] dt.

Shorter Alternative: Just check it for λ = i, this is the only new thing overthe real variable theory.5.

w (t) z (t) |βα =Z β

α

d

dt[w (t) z (t)] dt =

Z β

α

w(t)z(t)dt+

Z β

α

w(t)z(t)dt.

6. Let ρ ≥ 0 and θ ∈ R be chosen so thatZ β

α

w (t) dt = ρeiθ,

then ¯¯Z β

α

w (t) dt

¯¯ = ρ = e−iθ

Z β

α

w (t) dt =

Z β

α

e−iθw (t) dt

=

Z β

α

Re£e−iθw (t)

¤dt ≤

Z β

α

¯Re£e−iθw (t)

¤¯dt

≤Z β

α

¯e−iθw (t)

¯dt =

Z β

α

|w (t)| dt.


2.3. Application. We would like to use the above ideas to find a “natural” exten-sion of the function ex to a function ez with z ∈ C. The idea is that since

d

dtetx = xetx with e0x = 1

we might try to define ez so that

(2.1)d

dtetz = zetz with e0z = 1.

Proposition 2.4. If there is a function ez such that Eq. (2.1) holds, then itsatisfies:

(1) e−z = 1ez and

(2) ew+z = ewez.

Proof. 1. By the product rule,

d

dt

£e−tzetz

¤= −ze−tzetz + e−tzzetz = 0

and therefore, e−tzetz = e−0ze0z = 1. Taking t = 1 proves 1.2. Again by the product rule,

d

dt

he−t(w+z)etwetz

i= 0

and so e−t(w+z)etwetz = e−t(w+z)etwetz|t=0 = 1. Taking t = 1 then showse−(w+z)ewez = 1 and then using Item 1. proves Item 2.According to Proposition 2.4, to find the desired function ez it suffices to find

eiy. So let us writeeit = x (t) + iy(t)

then by assumption ddte

it = ieit with ei0 = 1 implies

x+ iy = i (x+ iy) = −y + ix with x(0) = 1 and y (0) = 0

or equivalently that

x = −y, y = x with x(0) = 1 and y (0) = 0.

This equation implies

x(t) = −y (t) = −x(t) with x (0) = 1 and x (0) = 0

which has the unique solution x (t) = cos t in which case y (t) = − ddt cos t = sin t.

This leads to the following definition.

Definition 2.5 (Euler’s Formula). For θ ∈ R let eiθ := cos θ + i sin θ and forz = x+ iy let

(2.2) ez = exeiy = ex (cos y + i sin y) .

Quickly review ez and its properties, in particular Euler’s formula.

Theorem 2.6. The function ez defined by Eq. (2.2) satisfies Eq. (2.1) and hencethe results of Proposition 2.4. Also notice that ez = ez.

8 BRUCE K. DRIVER

Proof. This is proved on p. 112 of the book and the proof goes as follows,d

dtetz =

d

dt

£etxeity

¤= xetxeity + etxiyeity = zetxeity = zetz.

The last equality follows from

ez = ex (cos y + i sin y) = ex(cos y + i sin y) = ex (cos y − i sin y)

= ex (cos (−y) + i sin (−y)) = ez.

Corollary 2.7 (Addition formulas). For α, β ∈ R we havecos (α+ β) = cosα cosβ − sinα sinβsin (α+ β) = cosα sinβ + cosβ sinα.

Proof. These follow by comparing the real and imaginary parts of the identity

eiαeiβ = ei(α+β) = cos (α+ β) + i sin (α+ β)

while

eiαeiβ = (cosα+ i sinα) · (cosβ + i sinβ)

= cosα cosβ − sinα sinβ + i (cosα sinβ + cosβ sinα) .


3. (10/1/03)

Exercise 3.1. Suppose a, b ∈ R, showZ T

0

eateibtdt =

Z T

0

e(a+ib)tdt =

Z T

0

1

a+ ib

d

dte(a+ib)tdt =

1

a+ ib

£eaT eibT − 1¤ .

By comparing the real and imaginary parts of both sides of this integral find explicitformulas for the two real integralsZ T

0

eat cos (bt) dt andZ T

0

eat sin (bt) dt.

3.1. Polar/Exponential Form of Complex Numbers: Sections 6 — 9. Bruce:Give the geometric interpretation of each of the following properties.

(1) z = reiθ = |z| eiθ.(2) z = |z| e−iθ and z−1 = z/ |z|2 = |z|−1 e−iθ(3) If w = |w| eiα then

zw = |z| |w| ei(θ+α) andz/w = zw−1 = |z| eiθ · |w|−1 e−iα = |z| |w|−1 ei(θ−α).

In particularzn = |z|n einθ for n ∈ Z.

Notation 3.2. If z 6= 0 we let θ = Arg (z) if −π < θ ≤ π and z = |z| eiθ while wedefine

arg (z) =©θ ∈ R : z = |z| eiθª .

Notice thatarg (z) = Arg (z) + 2πZ.

Similarly we define Log (z) = ln |z|+ iArg (z) and

log (z) = ln |z|+ i arg (z) = ln |z|+ iArg (z) + 2πiZ.

Example 3.3.(1) Work out (1 + i)

¡√3 + i

¢in polar form.

(1 + i)³√3 + i

´=√2eiπ/4 · 2eiπ/6 = 2

√2ei5π/12.

Note here that

arg (1 + i) = π/4 + 2πZ and arg³√3 + i

´= π/6 + 2πZ

Arg (1 + i) = π/4 and Arg³√3 + i

´= π/6

(2) Let α = tan−1 (1/2) then

5i

2 + i=

5eiπ/2√5ei tan−1(1/2)

=√5ei(π/2−tan

−1(1/2)) = 1 + 2i

by drawing the triangles.

10 BRUCE K. DRIVER

(3) General theory of finding nth — roots if a number z = ρeiα. Let w = reiθ

then z = ρeiα = wn = rneinθ happens iff

ρ = |z| = |wn| = |w|n = rn or r = ρ1/n and

eiα = einθ i.e. ei(nθ−α) = 1, i.e. nθ − α ∈ 2πZ.Therefore

z1/n = |z|1/n ei 1n (α+2πZ) = |z|1/n ei 1n arg(z)

=n|z|1/n ei 1n (α+2πk) : k = 0, 1, 2, . . . , n− 1

o.

(4) Find all fourth roots of (1 + i) .

(1 + i) =√2ei(π/4+2πZ)

and so

(1 + i)1/4 = 21/8ei(π/16+18πZ) =

n21/8ei(π/16+

12πk) : k = 0, 1, 2, 3

o.


4. (10/03/2003)

4.1. More on Roots and multi-valued arithmetic.

Notation 4.1. Suppose A ⊂ C and B ⊂ C, the we letA ·B := ab : a ∈ A and b ∈ B andA±B := a± b : a ∈ A and b ∈ B .

Proposition 4.2. arg (zw) = arg (z) + arg (w) while it is not in general true theArg (zw) = Arg (z) + Arg (w) .

Proof. Suppose z = |z| eiθ and w = |w| eiα, thenarg (zw) = (θ + α+ 2πZ)

whilearg (z) + arg (w) = (θ + 2πZ) + (α+ 2πZ) = (θ + α+ 2πZ) .

Example: Let z = i and w = −1, then Arg(i) = π/2 and Arg (−1) = π so that

Arg(i) + Arg (−1) = 3π

2

whileArg(i · (−1)) = −π/2.

The following proposition summarizes item 3. of Example 3.3 above and givesan application of Proposition 4.2.

Proposition 4.3. Suppose that w ∈ C, then the set of nth — roots, w1/n of w is

w1/n = np|w|ei 1n arg(w).

Moreover if z ∈ C then(4.1) (wz)1/n = w1/n · z1/nIn particular this implies if w0 is an nth— root of w, then

w1/n =nw0e

i kn2π : k = 0, 1, . . . , n− 1o.

DRAW picture of the placement of the roots on the circle of radius np|w|.

Proof. It only remains to prove Eq. (4.1) and this is done using

w1/n · z1/n = np|w|ei 1n arg(w) n

p|z|ei 1n arg(z)

= np|w| |z|ei 1n [arg(w)+arg(z)] = n

p|wz|ei 1n arg(wz)

= (wz)1/n

.

Theorem 4.4 (Quadratic Formula). Suppose a, b, c ∈ C with a 6= 0 then the generalsolution to the equation

az2 + bz + c = 0

is

z =−b± ¡b2 − 4ac¢1/2

2a.

12 BRUCE K. DRIVER

Proof. The proof goes as in the real case by observing

0 = az2 + bz + c = a

µz +

b

2a

¶2+ c− b2

4a

and so µz +

b

2a

¶2=

b2 − 4ac4a2

.

Taking square roots of this equation then shows

z +b

2a=

¡b2 − 4ac¢1/2

2awhich is the quadratic formula.

4.2. Regions and Domains:(1) Regions in the plain. Definition: a domain is a connected open subset

of C. Examples:(a) z : |z − 1 + 2i| < 4 .(b) z : |z − 1 + 2i| ≤ 4 .(c) z : |z − 1 + 2i| = 4(d)

©z : z = reiθ with r > 0 and − π < θ < π

ª(e)

©z : z = reiθ with r ≥ 0 and − π < θ ≤ π

ª.


5. (10/06/2003)

5.1. Functions from C to C..(1) Complex functions, f : D → C. Point out that f (z) = u (x, y) + iv (x, y)

where z = x+ iy ∈ D. Examples: (Mention domains)(a) f (z) = z, u = x, v = y(b) f (z) = z2, u = x2 − y2, v = 2xy also look at it as f

¡reiθ

¢= r2ei2θ.

(i) So rays through the origin go to rays through the origin.(ii) Also arcs of circles centered at 0 go over to arcs of circles centered

at 0.(iii) Also notice that if we hold x constant, then y = v/2x and so

u = x2 − v2

4x2 which is the graph of a parabola.(iv) Bruce !!: Do the examples where Re f (z) = 1 and Im f (z) = 1

to get pre-images which are two hyperbolas. Explain the orien-tation traversed. See Figure 1 below.

(c) f (z) = az, if a = reiθ, then f (z) scales z by r and then rotates by θdegrees. If a = α+ iβ, then u = αx− βy, v = αy = βx.

(d) f (z) = z, this is reflection about the x− axis.(e) f (z) = 1/z is inversion, notice that f

¡reiθ

¢= 1

reiθ= 1

r e−iθ, draw

picture.(f) f (z) = ez = ex+iy, u = ex cos y and v = ex sin y.

(i) Show what happens to the line x = 2 and the line y = π/4.

(g) f¡reiθ

¢= r

12 ei

12θ for −π < θ ≤ π. Somewhat painful to write u, v in

this case.

14 BRUCE K. DRIVER

Figure 1. Pre-images of lines for f (z) = z2.


16 BRUCE K. DRIVER

5.2. Continuity and Limits.

5.3. ε — Notation. In this section, U will be an open subset of C, f : U → C afunction and ε(z) will denote a generic function defined for z near zero such thatlimz→0 ε (z) = 0.

Definition 5.1. (1) limz→z0 f (z) = L iff f (z0 +∆z) = L+ ε (∆z)(2) f is continuous at z0 if limz→z0 f (z) = f (z0) = f (limz→z0 z) .(3) f is differentiable at z0 with derivative L iff

limz→z0

f (z)− f (z0)

z − z0= L

or equivalently iff

(5.1)f (z0 +∆z)− f (z0)

∆z=

f (z0 +∆z)− f (z0)

z0 +∆z − z0= L+ ε (∆z)

or equivalently,

(5.2) f (z0 +∆z)− f (z0) = (L+ ε (∆z))∆z.

Proposition 5.2. The functions f (z) := z, f (z) = Re z, and f (z) = Im z are allcontinuous functions which are not complex differentiable at any point z ∈ C. Thefollowing functions are complex differentiable at all points z ∈ C:

(1) f (z) = z with f 0 (z) = 1.(2) f (z) = 1

z with f 0 (z) = −z−2.(3) f(z) = ez with f 0 (z) = ez.

Proof. For the first assertion we have¯z0 +∆z − z0

¯= |∆z|→ 0

|Re (z0 +∆z)−Re z0| = |Re∆z| ≤ |∆z|→ 0 and

|Im (z0 +∆z)− Im z0| = |Im∆z| ≤ |∆z|→ 0.

For differentiability,f (z +∆z)− f (z)

∆z=∆z

∆zwhich has no limit as ∆z → 0. Indeed, consider what happens for ∆z = x and∆z = iy with x, y ∈ R and x, y → 0. Similarly

Re (z0 +∆z)−Re z0∆z

=Re∆z

∆z

as no limit as ∆z → 0.

(1)f (z +∆z)− f (z)

∆z= 1→ 1 as ∆z → 0.

(2) Let us first shows that 1/z is continuous, for this we have¯(z +∆z)

−1 − z−1¯=

¯z − (z +∆z)z (z +∆z)

¯=

¯1

z

¯ ¯1

z +∆z

¯|∆z|

≤¯1

z

¯ ¯1

|z|− |∆z|¯|∆z| ≤ 2

|z|2 |∆z|→ 0.


We now use this to compute the derivative,

f (z +∆z)− f (z)

∆z=(z +∆z)−1 − z−1

∆z

=1

z+∆z − 1z

∆z=

1

∆z

z − (z +∆z)z (z +∆z)

= − 1

z (z +∆z)→ − 1

z2.

where the continuity of 1/z was used in taking the limit.(3) Since

ez+∆z − ez

∆z= ez

e∆z − 1∆z

it suffices to showe∆z − 1∆z

→ 1 as ∆z → 0.

This follows from,

e∆z − 1∆z

=1

∆z

Z 1

0

d

dtet∆zdt =

1

∆z∆z

Z 1

0

et∆zdt =

Z 1

0

et∆zdt

which implies

e∆z − 1 = ∆zZ 1

0

et∆zdt = ε (∆z)

and therefore¯e∆z − 1∆z

− 1¯=

¯Z 1

0

£et∆z − 1¤ dt¯ ≤ Z 1

0

¯et∆z − 1¯ dt = Z 1

0

|ε (t∆z)| dt→ 0 as ∆z → 0.

Alternative 1.,Z 1

0

et∆zdt =

Z 1

0

et∆zd [t− 1]

=¡et∆z [t− 1]¢1

0−Z 1

0

d

dtet∆z [t− 1] dt

= 1−∆zZ 1

0

et∆z [t− 1] dtfrom which it should be clear that

e∆z − 1∆z

− 1 = (∆z) .

Alternative 2. Write ∆z = x + iy, then and use the definition of thereal derivative to learn

e∆z = ex+iy = ex (cos y + i sin y) =¡1 + x+O

¡x2¢¢ ¡

1 + iy +O¡y2¢¢

= 1 + x+ iy +O³|∆z|2

´= 1 +∆z +O

³|∆z|2

´.

18 BRUCE K. DRIVER

6. (10/08/2003 and 10/10/2003) Lectures 6-7

Go over the function f (z) = ez in a bit more detail than was done in class usingAlternative 2 above to show

(6.1) lim∆z→0

e∆z − 1∆z

= 1

To do this write ∆z = x + iy, then and use Taylor’s formula with remainder forreal functions to learn

e∆z = ex+iy = ex (cos y + i sin y) =¡1 + x+O

¡x2¢¢ ¡

1 + iy +O¡y2¢¢

= 1 + x+ iy +O³|∆z|2

´= 1 +∆z +O

³|∆z|2

´which implies Eq. (6.1).

Exercise 6.1. Suppose that f 0 (0) = 5 and g (z) = f (z) . Show g0 (0) does notexists.Solution:

g (z)− g (0)

z=

f (z)− f (0)

z=(5 + (z)) z

zand the latter does not have a limit by Proposition 5.2.

BRUCE: Do examples in this section before giving proofs.

Definition 6.2. Limits involving ∞,

(1) limz→∞ f (z) = w iff limz→0 f (1/z) = w.(2) limz→w f (z) =∞ iff limz→w

1f(z) = 0.

(3) limz→∞ f (z) =∞ iff limz→0 1f(1/z) = 0.

BRUCE: Explain the motivation via stereographic projection, see Figure 2.

Figure 2. The picture behind the limits at infinity.

Theorem 6.3. If limz→z0 f (z) = L and limz→z0 g (z) = K then(1) limz→z0 [f (z) + g (z)] = L+K.(2) limz→z0 [f (z) g (z)] = LK(3) If z → h (z) = f (g (z)) is continuous at z0 if g is continuous at z0 and f

is continuous at w0 = g (z0) .

(4) limz→z0

hf(z)g(z)

i= L

K provided K 6= 0.


(5) We also have limz→z0 f (z) = L iff limz→z0 Re f (z) = ReL andlimz→z0 Im f (z) = ImL.

Proof.(1)

f (z0 +∆z) + g (z0 +∆z) = L+ ε (∆z) +K + ε (∆z) = (L+K) + ε (∆z) .

(2)

f (z0 +∆z) · g (z0 +∆z) = [L+ ε (∆z)] · [K + ε (∆z)]

= LK +Kε (∆z) + Lε (∆z) + ε (∆z) ε (∆z) = LK + ε (∆z) .

(3) Well,

h (z0 +∆z)− h (z0) = f (g (z0 +∆z))− f (g (z0))

= f (g (z0) + ε (∆z))− f (g (z0)) = ε (ε (∆z))→ 0 as ∆z → 0.

(4) This follows directly using

f (z0 +∆z)

g (z0 +∆z)− L

K=

L+ ε (∆z)

K + ε (∆z)− L

K=(L+ ε (∆z))K − L (K + ε (∆z))

K2 +Kε (∆z)

=ε (∆z)

K2 + ε (∆z)= ε (∆z) .

or more simply using item 3. and the fact 1/z is continuous so that

limz→z0

h1

g(z)

i= 1

K .

(5) This follows from item 1. and the continuity of the functions z → Re z andz → Im z.

Theorem 6.4. If f 0 (z0) = L and g0 (z0) = K then

(1) f is continuous at z0,(2) d

dz [f (z) + g (z)] |z=z0 = L+K

(3) ddz [f (z) g (z)] |z=z0 = f 0 (z0) g (z0) + f (z0) g

0 (z0)(4) If w0 = f (z0) and g0 (w0) exists then h(z) := g(f (z)) is differentiable as

z0 andh0 (z0) = g0 (f (z0)) f 0 (z0) .

(5)d

dz

·f (z)

g (z)

¸=

f 0g − g0fg2

|z=z0 .

(6) If z (t) is a differentiable curve, then ddtf (z (t)) = f 0 (z (t)) z (t) .

Proof. To simplify notation, let ∆f = f (z +∆z)−f (z) and ∆g = g (z +∆z)−g (z) and recall that recall that ∆f → 0 and ∆g → 0 as ∆z → 0, i.e. ∆f = ε (∆z) .

(1) This follows from Eq. (5.2).(2)

[f (z +∆z) + g (z +∆z)]− [f (z) + g (z)]

∆z=∆f

∆z+∆g

∆z→ f 0 (z) + g0 (z)

20 BRUCE K. DRIVER

(3)

f (z +∆z) g (z +∆z)− f (z) g (z)

∆z=(f (z) +∆f) (g (z) +∆g)− f (z) g (z)

∆z

=f (z)∆g +∆fg (z) +∆f∆g

∆z

= f (z)∆g

∆z+ g (z)

∆f

∆z+∆g

∆z∆f → f (z) g0 (z) + g (z) f 0 (z) .

(4) Recall that ∆f = ε (∆z) and so

∆h := h(z +∆z)− h (z) = g(f (z +∆z))− g(f (z))

= g (f (z) +∆f)− g (f (z))

= [g0 (f (z)) + ε (∆f)]∆f = [g0 (f (z)) + ε (∆z)]∆f.

Therefore∆h

∆z= [g0 (f (z)) + ε (∆z)]∆f → g0 (f (z)) f 0 (z) .

(5) This follows from the product rule, the chain rule and the fact that ddz z−1 =

−z−2.(6) In order to verify this item, we first need to observe that z (t) exists iff

lim∆t→0z(t+∆t)−z(t)

∆t . Recall that we defined

z (t) =d

dtRe z (t) + i

d

dtIm z (t) .

Since the limit of a sum is a sum of a limit if ddt Re z (t) and

ddt Im z (t) exist

then lim∆t→0z(t+∆t)−z(t)

∆t exists. Conversely if w = lim∆t→0z(t+∆t)−z(t)

∆texists, then

lim∆t→0

¯Re z (t+∆t)− Re z (t)

∆t−Rew

¯= lim∆t→0

¯Re

µz (t+∆t)− z (t)

∆t− w

¶¯≤ lim∆t→0

¯z (t+∆t)− z (t)

∆t− w

¯= 0

which shows ddt Re z (t) exists. Similarly one shows

ddt Im z (t) exists as well.

Now for the proof of the chain rule: let ∆z := z (t+∆t)− z (t)

f (z (t+∆t))− f (z (t))

∆t=[f 0 (z (t)) + ε (∆z)]∆z

∆t

= [f 0 (z (t)) + ε (∆z)]∆z

∆t→ f 0 (z (t)) z (t) .

Example 6.5.(1) z is continuous, z, Re z, Im z are continuous and polynomials in these vari-

ables.(2) limz→z0 z

n = zn0 , Proof by induction.(3) limz→1 z2−1

z−1 = limz→1 (z + 1) = 2.(4) limz→1 1

z3−1 =∞, where by definition limz→z0 f (z) =∞ iff limz→z01

f(z) =

0.(5) limz→∞ z2+1

z2−1 = 1 where by definition limz→∞ f (z) = L iff limz→0 f¡1z

¢=

L.


(6) limz→∞ z25+1z24−z6+7z2−5 = ∞ where by definition limz→∞ f (z) = ∞ iff

limz→0 1f( 1z )

= 0.

(7) ez is continuous, proof

ez0+∆z − ez0 = ez0¡e∆z − 1¢ = ∆zez0 Z 1

0

et∆zdt = ε (∆z)

(8) limz→−1 z+1z2+(1−i)z−i , since z

2+(1− i) z− i = 1− (1− i)− i = 0 at z = −1we have to factor the denominator. By the quadratic formula we have

z =− (1− i)±

q(1− i)2 + 4i

2=− (1− i)±

q(1 + i)2

2

=− (1− i)± (1 + i)

2= i,−1

and thusz2 + (1− i) z − i = (z − i) (z + 1)

and we thus have

limz→−1

z + 1

z2 + (1− i) z − i= lim

z→−1z + 1

(z − i) (z + 1)=

1

−1− i= −1− i

2.

Example 6.6. Describe lots of analytic functions and compute their derivatives:for example z2, p (z) , ez

2

, e1/z, sin (z) cos (z) , etc.

Example 6.7 (Important Example).Z 1

0

(1 + it)3dt =

1

4i(1 + it)

4 |10 =1

4i

h(1 + i)

4 − 1i=5

4i.

If we did this the old fashion way it would be done as followsZ 1

0

(1 + it)3 dt =

Z 1

0

£1 + 3it− 3t2 − it3

¤dt = 1− 1 + i

µ3

2− 14

¶=5

4i.

Example 6.8.Z π/2

0

e(1+i)π sin t cos tdt =1

π

e(1+i)π sin t

1 + i=1

π

1

1 + i

heπ(1+i) − 1

i=1

π

1

1 + i[−eπ − 1] .

22 BRUCE K. DRIVER

7. Study Guide for Math 120A Midterm 1 (Friday October 17, 2003)

(1) C := z = x+ iy : x, y ∈ R with i2 = −1 and z = x − iy. The complexnumbers behave much like the real numbers. In particular the quadraticformula holds.

(2) |z| =px2 + y2 =

√zz, |zw| = |z| |w| , |z + w| ≤ |z| + |w| , Re z = z+z

2 ,

Im z = z−z2i , |Re z| ≤ |z| and |Im z| ≤ |z| . We also have zw = zw and

z + w = z + w and z−1 = z|z|2 .

(3) z : |z − z0| = ρ is a circle of radius ρ centered at z0.z : |z − z0| < ρ is the open disk of radius ρ centered at z0.z : |z − z0| ≥ ρ is every thing outside of the open disk of radius ρ cen-

tered at z0.(4) ez = ex (cos y + i sin y) , every z = |z| eiθ.(5) arg (z) =

©θ ∈ R : z = |z| eiθª and Arg (z) = θ if −π < θ ≤ π and z =

|z| eiθ. Notice that z = |z| ei arg(z)(6) z1/n = n

p|z|ei arg(z)n .(7) limz→z0 f (z) = L. Usual limit rules hold from real variables.(8) Mapping properties of simple complex functions(9) The definition of complex differentiable f (z) . Examples, p (z) , ez, ep(z),

1/z, 1/p (z) etc.(10) Key points of ez are is d

dz ez = ez and ezew = ez+w.

(11) All of the usual derivative formulas hold, in particular product, sum, andchain rules:

d

dzf (g (z)) = f 0 (g (z)) g0 (z)

andd

dtf (z (t)) = f 0 (z (t)) z (t) .

(12) Re z, Im z, z, are nice functions from the real - variables point of view butare not complex differentiable.

(13) Integration: Z b

a

z (t) dt :=

Z b

a

x (t) dt+ i

Z b

a

y (t) dt.

All of the usual integration rules hold, like the fundamental theorem ofcalculus, linearity and integration by parts.


8. (10/13/2003) Lecture 8

Definition 8.1 (Analytic and entire functions). A function f : D → C is said tobe analytic (or holomorphic) on an open subset D ⊂ C if f 0 (z) exists for allz ∈ D. An analytic function f on C is said to be entire.

8.1. Cauchy Riemann Equations in Cartesian Coordinates. If f (z) is com-plex differentiable, then by the chain rule

∂xf (x+ iy) = f 0 (x+ iy) while

∂yf (x+ iy) = if 0 (x+ iy) .

So in order for f (z) to be complex differentiable at z = x+ iy we must have

(8.1) fy (x+ iy) := ∂yf (x+ iy) = i∂xf (x+ iy) = ifx (x+ iy) .

Writing f = u + iv, Eq. (8.1) is equivalent to uy + ivy = i (ux + ivx) and thusequivalent to

(8.2) uy = −vx and ux = vy.

Theorem 8.2 (Cauchy Riemann Equations). Suppose f (z) is a complex function.If f 0 (z) exists then fx (z) and fy (z) exists and satisfy Eq. (8.1), i.e.

∂yf (z) = i∂xf (z)

Conversely if fx and fy exists and are continuous in a neighborhood of z, then f 0 (z)exists iff Eq. (8.1) holds.

Proof. (I never got around to giving this proof.) We have already proved thefirst part of the theorem. So now suppose that fx and fy exists and are continuousin a neighborhood of z and Eq. (8.1) holds. To simplify notation let us supposethat z = 0 and ∆z = x+ iy, then

f (x+ iy)− f (0) = f (x+ iy)− f (x) + f (x)− f (0)

=

Z 1

0

d

dtf (x+ ity) dt+

Z 1

0

d

dtf (tx) dt

=

Z 1

0

[yfy (x+ ity) + xfx (tx)] dt

=

Z 1

0

[iyfx (x+ ity) + xfx (tx)] dt

=

Z 1

0

[iy (fx (x+ ity)− fx (0)) + x (fx (tx)− fx (0)) + fx (0) (x+ iy)] dt

= zfx (0) +

Z 1

0

[iy (fx (x+ ity)− fx (0)) + x (fx (tx)− fx (0))] dt

= zfx (0) +

Z 1

0

[iyε (z) + xε (z)] dt = zfx (0) + |z| ε (|z|) .

Fact 8.3 (Amazing Fact). We we will eventually show, that if f is analytic on anopen subset D ⊂ C, then f is infinitely complex differentiable on D, i.e. f analyticimplies f 0 is analytic!!! Note well: it is important that D is open here. SeeRemark 8.6 below.

24 BRUCE K. DRIVER

Example 8.4. Consider the following functions:(1) f (z) = x + iby. In this case fx = 1 while fy = ib so fy = ifx iff b = 1. In

this case f (z) = z.(2) f (z) = z2, then u = x2 − y2 and v = 2xy, uy = −2y = −vx and vy = 2x =

ux, which shows that f (z) = z2 is complex differentiable.(3) f (z) = ez = ex (cos y + i sin y) , so fx = f while

fy = eix (− sin y + i cos y) = if = ifx

which again shows that f is complex differentiable.(4) Also work out the example f (z) = 1/z = x−iy

x2+y2 ,

fx =x2 + y2 − 2x (x− iy)

(x2 + y2)2=

y2 − x2 + 2ixy

|z|4Note

−µ1

z

¶2= − (x− iy)

2

(x2 + y2)2= −x

2 − y2 − 2ixy(x2 + y2)2

=y2 − x2 + 2ixy

|z|4 = fx.

Similarly

fy =−i ¡x2 + y2

¢− 2y (x− iy)

(x2 + y2)2=−ix2 + iy2 − 2yx

(x2 + y2)2= ifx

and all of this together shows that f 0 (z) = − 1z2 for z 6= 0.

Corollary 8.5. Suppose that f = u+ iv is complex differentiable in an open set D,then u and v are harmonic functions, i.e. that real and imaginary parts of analyticfunctions are harmonic.

Proof. The C.R. equations state that vy = ux and vx = −uy, thereforevyy = uxy = uyx = −vxx.

A similar computation works for u.

Remark 8.6. The only harmonic functions f : R→ R are straight lines, i.e. f (x) =ax + b. In particular, any harmonic function f : R→ R is infinitely differentiable.This should shed a little light on the Amazing Fact in Example 8.3.

Example 8.7 (The need for continuity in Theorem 8.2). Exercise 6, on p. 69.Consider the function

f (z) =

½z2

z if z 6= 00 if z = 0.

Then

fx (0) = limx→0

f (x)− f (0)

x= lim

x→0x

x= 1

while

fy (0) = limy→0

f (iy)− f (0)

y= lim

y→0−y2/iy

y= −1

i= i = ifx (0) .

Thus the Cauchy Riemann equations hold at 0. However,

f 0 (0) = limz→0

f (z)− f (0)

z= lim

z→0z2

z2


does not exist. For example taking z = x real and z = xeiπ/4 we get

limz=x→0

z2

z2= 1 while lim

z=xeiπ/4→0z2

z2= lim

z=xeiπ/4→0−ix2ix2

= −1.

26 BRUCE K. DRIVER

9. (10/15/2003) Lecture 9

Example 9.1. Show Re f and Im f are harmonic when f (z) = z2 and f (z) = ez.

Definition 9.2. A function f : D→ C is analytic on an open set D iff f 0 (z) iscomplex differentiable at all points z ∈ D.

Definition 9.3. For z 6= 0, let log z = w ∈ C : ew = z .Writing z = |z| eiθ we and w = x + iy, we must have |z| eiθ = exeiy and this

implies that x = ln |z| and y = θ + 2πn for some n. Therefore

log z = ln |z|+ i arg z.

Definition 9.4. Log(z) = ln |z| + iArg(z), so Log(reiθ) = ln r + iθ if r > 0 and−π < θ ≤ π. Note this function is discontinuous at points z where Arg(z) = π.

Definition 9.5. Given a multi-valued function f : D → C, we say a F : D0 ⊂D → C is a branch of f if F (z) ∈ f (z) for all z ∈ D0 and F is continuous on D0.Here D0 is taken to be an open subset of D.

Example 9.6 (A branch of log (z) : a new analytic function). A branch of log(z).Here we take D = z = x+ iy : x > 0 .

f (z) = Log(z) =1

2ln¡x2 + y2

¢+ i tan−1 (y/x) .

Recall that ddt tan

−1 (t) = 1t2+1 so we learn

fx =1

2

2x

x2 + y2+ i

− yx2

1 + (y/x)2 =

x

x2 + y2− i

y

x2 + y2=

1

|z|2 z =1

z

fy =1

2

2y

x2 + y2+ i

1x

1 + (y/x)2=

y

x2 + y2+ i

x

x2 + y2= i

1

z= ifx

from which it follows that f is complex differentiable and f 0 (z) = 1z .

Note that for Im z > 0, we have Log(z) = f¡1i z¢+ iπ/2 which shows Log(z) is

complex differentiable for Im z > 0.Similarly, if Im z < 0, we have Log(z) = f (iz) − iπ/2 which shows Log(z) is

complex differentiable for Im z < 0.Combining these remarks shows that Log(z) is complex differentiable on C \

(−∞, 0].

Example 9.7 (Homework Problem: Problem 7a on p.74). Suppose that f is acomplex differentiable function such that Im f = 0. Then fx and fy are real andfy = ifx can happen iff fx = fy = 0. But this implies that f is constant.

Example 9.8 (Problem 7b on p.74 in class!). Now suppose that |f (z)| = c 6= 0for all z is a domain D. Then

f (z) =|f (z)|2f (z)

=c2

f (z)

which shows f is complex differentiable and from this it follows that Re f = f+f2

and Im f = f−f2i are real valued complex differentiable functions. So by the previous

example, both Re f and Im f are constant and hence f is constant.

Test #1 was on 10/17/03. This would have been lecture 10.


10. (10/20/2003) Lecture 10

Definition 10.1 (Analytic Functions). A function f : D→ C is said to be analytic(or holomorphic) on an open subset D ⊂ C if f 0 (z) exists for all z ∈ D.

Proposition 10.2. Let f = u+ iv be complex differentiable, and suppose the levelcurves u = a and v = b cross at a point z0 where f 0 (z0) 6= 0 then they cross at aright angle.

Proof. The normals to the level curves are given by ∇u and ∇v, so it sufficesto observe from the Cauchy Riemann equations that

∇u ·∇v = uxvx + uyvy = vyvx + (−vx) vy = 0.Draw Picture.Alternatively: Parametrize u = a and v = b by z (t) and w (t) so that z (0) =

z0 = w (0) . Then f (z (t)) = a+ iv (z (t)) and f (w (t)) = u (w (t)) + ib and

iβ =d

dt|0f (z (t)) = f 0 (z0) z (0) while

α =d

dt|0f (w (t)) = f 0 (z0) w (0)

where α = ddt |0u (w (t)) and β = d

dt |0v (z (t)) . Therefore

Rehz (0) w (0)

i= Re

·iβ

f 0 (z0)α

f 0 (z0)

¸= 0.

Alternatively,

∇u ·∇v = uxvx + uyvy = vyvx + (−vx) vy = 0.

Example 10.3 (Trivial case).

543210-1-2-3-4-5

5

4

3

2

10

-1

-2

-3

-4

-5

x

y

x

y

Some Level curves of Re f and Im f for f (z) = z.

28 BRUCE K. DRIVER

Example 10.4 (Homework).

543210-1-2-3-4-5

5

4

3

2

10

-1

-2

-3

-4

-5

x

y

x

y

Some Level curves of Re f and Im f for f (z) = z2.

10.1. Harmonic Conjugates.

Definition 10.5. Given a harmonic function u on a domain D ⊂ C, we say v is aharmonic conjugate to u if v is harmonic and u and v satisfy the C.R. equations.

Notice that v is uniquely determined up to a constant since if w is anotherharmonic conjugate we must have

wy = ux = vy and wx = −uy = vx.

Therefore ddtw (z (t)) =

ddtv (z (t)) for all paths z in D and hence w = v + C on D.

Proposition 10.6. f = u+ iv is complex analytic on D iff u and v are harmonicconjugates.

Example 10.7. Suppose u (x, y) = x2 − y2 we wish to find a harmonic conjugate.For this we use

vy = ux = 2x and

vx = −uy = 2yto conclude that v = 2xy + C (x) and then 2y = vx = 2y + C 0 (x) which impliesC 0 (x) = 0 and so C =const. Thus we find

f = u+ iv = x2 − y2 + i2xy + iC = z2 + iC

is analytic.

Example 10.8. Now suppose that u = 2xy. In this case we have

vy = ux = 2y and

vx = −uy = −2x


and so v = y2

2 + C (x) and so −2x = vx = C 0 (x) from which we learn thatC (x) = −x2 + k. Thus we find

f = 2xy + i¡y2 − x2

¢+ ik = −iz2 + ik

is complex analytic.

Recall the following definitions:

Definition 10.9. For z 6= 0, let log z = w ∈ C : ew = z .Writing z = |z| eiθ we and w = x + iy, we must have |z| eiθ = exeiy and this

implies that x = ln |z| and y = θ + 2πn for some n. Therefore

log z = ln |z|+ i arg z.

Definition 10.10. Log(z) = ln |z|+ iArg(z), so Log(reiθ) = ln r + iθ if r > 0 and−π < θ ≤ π. Note this function is discontinuous at points z where Arg(z) = π.

Example 10.11. Find log 1, log i, log¡−1−√3i¢ .

Theorem 10.12 (Converse Chain Rule: Optional). Suppose f : D ⊂o C→U ⊂o Cand g : U ⊂o C→ C are functions such that f is continuous, g is analytic andh := g f is analytic, then f is analytic on the set D \z : g0(f(z)) = 0 . Moreoverf 0(z) = h0(z)/g0(f(z)) when z ∈ D and g0(f(z)) 6= 0.Proof. Suppose that z ∈ D and g0(f(z)) 6= 0. Let ∆f = f (z +∆z)− f(z) and

notice that ∆f = ε (∆z) because f is continuous at z. On one hand

h (z +∆z) = h(z) + (h0(z) + ε (∆z))∆z

while on the other

h (z +∆z) = g(f (z +∆z)) = g(f (z) +∆f)

= g(f(z)) + [g0(f(z) + ε (∆f)]∆f

= h(z) + [g0(f(z) + ε (∆z)]∆f.

Comparing these two equations implies that

(10.1) (h0(z) + ε (∆z))∆z = [g0 (f(z)) + ε (∆z)]∆f

and since g0(f(z)) 6= 0 we may conclude that∆f

∆z=

h0(z) + ε (∆z)

g0 (f(z)) + ε (∆z)→ h0(z)

g0 (f(z))as ∆z → 0,

i.e. f 0 (z) exists and f 0 (z) = h0(z)g0(f(z)) .

Definition 10.13 (Inverse Functions). Given a function f : C→ C we letf−1 (w) := z ∈ C : f (z) = w . In general this is a multivalued function and wewill have to choose a branch when we need an honest function.

Example 10.14. Since eLog(z) = z and Log(z) is continuous on D := C \ (−∞, 0],Log(z) is complex analytic on D and

1 =d

dzz =

d

dzeLog(z) = eLog(z)

d

dzLog(z) = z

d

dzLog(z),

i.e. we haved

dzLog(z) =

1

z.

30 BRUCE K. DRIVER

11. (10/22/2003) Lecture 11

Example 11.1. In fact the above example generalizes, suppose (z) is any branchof log (z) , that is is a continuous function on an open set D ⊂ C such thate (z) = z, then 0 (z) = 1/z. Indeed, this follows just as above using the converse tothe chain rule.

• Give the proof of Theorem 10.12.

Lemma 11.2. The following properties of log hold.(1) elog z = z(2) log ez = z + i2πZ(3) zn = en log(z) = elog z+log z+···+log z (n — times.)(4) z1/n = e

1n log z

(5) log z±1/n = ± 1n log z but be careful:

(6) log zn 6= n log z(7) log (wz) = logw + log z and in particular

log zn =

n — timesz | log z + log z + · · ·+ log z.

Proof.(1) This is by definition.(2) log ez = log ez+i2πZ = x+ i (y + 2πZ) = z + i2πZ.(3) If z = reiθ, then zn = rneinθ for any θ ∈ arg z, therefore

zn = rnein arg(z) = en ln rein arg(z) = en log z.

Better proof, if w ∈ log z, then z = ew so that zn = enw for any w ∈ log z,so zn = en log(z).

(4) We know

z1/n = |z|1/n ei 1n arg z = e1n ln|z|ei

1n arg z = e

1n log z.

(5) Now

log z±1/n = ln³|z|1/n

´± i

1

narg z + i2πZ = ln

³|z|1/n

´± i

1

narg z = ± 1

nlog z.

(6) On the other hand if z = |z| eiθ, thenlog zn = ln |z|n + i arg (zn) = n ln |z|+ i (nθ + 2πZ) = n ln |z|+ inθ + i2πZ

while

n log z = n (ln |z|+ iθ + i2πZ) = n ln |z|+ inθ + i2πnZ.

(7) This follows from the corresponding property arg (wz) and for ln,

log (wz) = ln |wz|+ i arg (wz) = ln |w|+ ln |z|+ i [arg (w) + arg (z)] = logw+ log z.

Definition 11.3. For c ∈ C, let zc := ec log z.

As an example let us work out ii :

ii = ei log i = eii(π/2+n2π) = e−(π/2+n2π).


Example 11.4. Let be a branch of log (z) , i.e. a continuous choice : D → Csuch that (z) ∈ log (z) for all z ∈ D then we define

d

dzzc =

d

dzec (z) = ec (z)c 0 (z)

= cec (z)1

z= cec (z)e− (z) = ce(c−1) (z) = czc−1.

The book writes P.V.zc = zc+ := zcLog := ecLog(z) for the principal value choice.Note with these definitions we have

z−c = e−c (z) =1

ec (z)=1

zc.

and when n ∈ N, then(zc)

n= encLog(z) = znc

however¡zcLog

¢dLog

=³ecLog(z)

´dLog

= edLog(ecLog(z)) = ed(cLog(z)+2πin)) = zdcLoge

i2πnd

for some integer n.

Definition 11.5 (Trig. and Hyperbolic Trig. functions:).

sin (z) :=eiz − e−iz

2i

cos (z) :=eiz + e−iz

2

tan (z) =sin (z)

cos (z)= −ie

iz − e−iz

eiz + e−iz

sinh (z) :=ez − e−z

2

cosh (z) :=ez + e−z

2

tanh (z) =sinh (z)

cosh (z)=

ez − e−z

ez + e−z.

Example 11.6. Basic properties of Trig. functions.(1) d

dz sin z = cos z andddz sinh z = cosh z

(2) ddz cos z = − sin z and d

dz cosh z = sinh z(3) sin z = −i sinh (iz) or sin iz = −i sinh (iiz) = −i sinh (−z) , i.e.

sin iz = i sinh z.

Alternatively

sin iz =eiiz − e−iiz

2i= −e

z − e−z

2i= i sinh z.

(4) cos z = cosh (iz) or cosh (z) = cos (iz) .(5) All the usual identities hold. For example

cos (w + z) = cosw cos z − sinw sin z(11.1)

sin (w + z) = sinw cos z + cosw sin z.(11.2)

32 BRUCE K. DRIVER

Indeed,

cosw cos z − sinw sin z = eiw + e−iw

2

eiz + e−iz

2− eiw − e−iw

2i

eiz − e−iz

2i

=1

4

h2ei(w+z) + 2e−i(w+z)

i= cos (w + z)

and (this one is homework)

sinw cos z + cosw sin z =eiw − e−iw

2i

eiz + e−iz

2+

eiw + e−iw

2

eiz − e−iz

2i

=1

4i

h2ei(w+z) − 2e−i(w+z)

i= sin (w + z) .

(6) In particular we have

cos (x+ iy) = cosx cosh y − i sinx sinh y(11.3)

sin (x+ iy) = sinx cosh y + i cosx sinh y.(11.4)

Indeed,

cos (x+ iy) = cosx cos iy − sinx sin iy= cosx cosh y − i sinx sinh y

and

sin (x+ iy) = sinx cos iy + cosx sin iy

= sinx cosh y + i cosx sinh y.


12. (10/24/2003) Lecture 12

Remark 12.1 (Roots Remarks).

(1) Warning: 1i = ei log 1 = ei(i2πZ) =©1, e±2π, e±4π, . . .

ª 6= 1(2) ii = ei log i = eii(π/2+n2π) = e−(π/2+n2π) = e−π/2

©1, e±2π, e±4π, . . .

ª.

(3) On the positive side we do have¡w2z

¢1/2= wz1/2 or more generally that

(wnz)1/n

= wz1/n

for any integer n. To prove this, ξ ∈ (wnz)1/n iff ξn = wnz iff

³ξw

ń= z iff

ξw ∈ z1/n iff ξ ∈ wz1/n.Alternatively,¡

w2z¢1/2

= e12 log(w

2z) = e12 [log(z)+log(w

2)] = e12 log(z)e

12 log(w

2) = z1/2e12 log(w

2).

Now if w = reiθ, then

logw2 = 2 ln r + i (2θ + 2πZ)and therefore,

e12 logw

2

= e[ln r+i(θ+πZ)] = ±reiθ = ±w.But ±wz1/2 = z1/2.

(4) log z1/2 = 12 log z. Indeed,

log z1/2 = log³e12 [ln|z|+i arg z]

´=1

2[ln |z|+ i arg z]+i2πZ =

1

2[ln |z|+ i arg z] =

1

2log z.

Example 12.2. Continuing Example 11.6 above.(1)

sin2 z + cos2 z =

·eiz − e−iz

2i

¸2+

·eiz + e−iz

2

¸2=1

44 = 1.

(2) Taking w = x and z = iy in the above equations shows

cos z = cosx cos iy − sinx sin iy= cosx cosh y − i sinx sinh y

and

sin z = sinx cos iy + cosx sin iy

= sinx cosh y + i cosx sinh y

(3) From this it follows that sin z = 0 iff sinx cosh y = 0 and cosx sinh y = 0.Since, cosh y is never zero we must have sinx = 0 in which case cosx 6= 0so that sinh y = 0 i.e. y = 0. So the only solutions to sin z = 0 happenwhen z is real and hence z = πZ. A similar argument works for cos z.

(4) Lets find all the roots of sin z = 2,

2 = sin z = sinx cosh y + i cosx sinh y

and socosx sinh y = 0 and sinx cosh y = 2.

Hence either y = 0 and sinx = 2 which is impossible of cosx = 0, i.e.x = π

2 +nπ for some integer n, and in this case sinx = (−1)n and we musthave (−1)n cosh y = 2 which can happen only for even n. Now cosh y = 2

34 BRUCE K. DRIVER

210-1-2

5

3.75

2.5

1.25

0

yy

Finding the roots of cosh y = 2 graphically.

iff (with ξ = ey) 2 = ξ+ξ−12 , i.e.

ξ2 + 1− 4ξ = 0or

ξ =4±√16− 4

2= 2±√4− 1 = 2±

√3.

Therefore y = ln¡2±√3¢ and we have

(12.1) sin z = 2 iff z =π

2+ 2nπ + i ln

³2±√3´for some n ∈ Z.

It should be noted that³2 +√3´³2−√3´= 4− 3 = 1

so that the previous equation may be written as

z =π

2+ 2nπ ± i ln

³2 +√3´.

Theorem 12.3. The inverse trig. functions

sin−1 (z) = −i log³iz +

¡1− z2

¢1/2ćos−1 (z) = −i log

³z + i

¡1− z2

¢1/2´tan−1 (z) =

i

2log

µi+ z

i− z

¶.

.


Moreover we haved

dzsin−1 (z) =

1√1− z2

d

dzcos−1 (z) =

−1√1− z2

d

dztan−1 (z) =

1

1 + z2

with appropriate choices of branches being specified.

Example 12.4.

cos−1 (0) =1

ilog (±i) = 1

ii³±π

2+ 2πZ

´=

½±π

2,±3π

2,±5π

2

¾so the zeros of the complex cosine function are precisely the zeros of the real cosinefunction. Similarly

sin−1 (2) = −i log³i2 + (1− 4)1/2

´= −i log

³i³2±√3´´

= −ihlog i+ log

³2±√3í= −i

hπ2+ 2πn+ ln

³2±√3í

= −ihi³π2+ 2πn

´+ ln

³2±√3í

=π

2+ 2πn− i ln

³2±√3´=

π

2+ 2πn± i ln

³2 +√3´

as before.

Proof.• cos−1 (w) : We have z ∈ cos−1 (w) iff

w = cos (z) =eiz + e−iz

2=

ξ + ξ−1

2

where ξ = eiz. Thusξ2 − 2wξ + 1 = 0

or

ξ =2w +

¡4w2 − 4¢1/22

= w +¡w2 − 1¢1/2

and therefore

iz = log ξ = log³w +

¡w2 − 1¢1/2´

and we have shown

cos−1 (w) = −i log³w +

¡w2 − 1¢1/2´

= −i log³w + i

¡1− w2

¢1/2´.

36 BRUCE K. DRIVER

13. (10/31/2003) Lecture 13 (Contour Integrals)

Lost two Lectures because of the big fire!!(Here I only computed d

dz tan−1 (z) in the proof below.)

Proof. Continuation of the proof.Let us now compute the derivative of this cos−1 (z) . For this we will need to

take a branch of f (z) of cos−1 z, say

(13.1) f (z) = −i ¡z + iQ¡1− z2

¢¢where is a branch of log and Q is a branch of the square-root. Then cos f (z) = zand differentiating this equations gives, − sin f (z) · f 0 (z) = 1 or equivalently that

f 0 (z) =1

− sin f (z) ∈ −1

(1− z2)1/2.

since sin f (z) ∈ ¡1− z2¢1/2

. The question now becomes which branch do we take.To determine this let us differentiate Eq. (13.1);

f 0 (z) =−i

z + iQ (1− z2)

½1− i

2Q (1− z2)(2z)

¾=

−iz + iQ (1− z2)

(Q¡1− z2

¢− iz

Q (1− z2)

)

=−1

z + iQ (1− z2)

(z + iQ

¡1− z2

¢Q (1− z2)

)=

−1Q (1− z2)

so we must use the same branch of the square-root used in Eq. (13.1). Hence wehave shown

“d

dzcos−1 (z) =

−1√1− z2

, ”

with the branch conditions determined as above.

• tan−1 (w) : We have z ∈ tan−1 (w) iff

w = tan (z) = −ieiz − e−iz

eiz + e−iz= −iξ − ξ−1

ξ + ξ−1= −iξ

2 − 1ξ2 + 1

where ξ = eiz. Thus¡ξ2 + 1

¢w + i

¡ξ2 − 1¢ = 0

orξ2 (w + i) = i− w

that is

ξ =

µi− w

i+ w

¶ 12

and hence

iz = log ξ = log

µi− w

i+ w

¶ 12

= −12log

µi+ w

i− w

¶so that

tan−1 (w) =i

2log

µi+ w

i− w

¶.


We have used here that log³η−

1n

´= − 1

n log η which happens because

n is an integer, see Lemma 11.2. Let us now compute the derivative oftan−1 (w) . In order to do this, let be a branch of log, and the f (w) =i2

³i+wi−w

´be a Branch of tan−1 (w) , then

d

dwf (w) =

i

2

1i+wi−w

d

dw

i+ w

i− w=

i

2

i− w

i+ w

(i− w) + (i+ w)

(i− w)2

= − 1

(i+ w) (i− w)=

1

(w + i) (w − i)=

1

1 + w2.

Thus we haved

dwtan−1 (w) =

1

1 + w2

where the formula is valid for any branch of tan−1 (w) that we have chosen.• sin−1 (w) : (This is done in the book so do not do in class.) We havez ∈ sin−1 (w) iff

w = sin (z) =eiz − e−iz

2i=

ξ − ξ−1

2i

where ξ = eiz. Thus

ξ2 − 1− 2iwξ = 0or

ξ =2iw +

¡−4w2 + 4¢1/22

= iw +¡1− w2

¢1/2and therefore

iz = log ξ = log³iw +

¡1− w2

¢1/2ánd we have shown

sin−1 (w) = −i log³iw +

¡1− w2

¢1/2´.

Example if w = 0, we have

sin−1 (0) =1

ilog (±1) = 1

iiπZ = πZ.

Suppose that

(13.2) f (w) = −i ¡iw +Q¡1− w2

¢¢where is a branch of log and Q is a branch of the square-root,then sin f (w) = w and so differentiating this equation in w givescos f (w) f 0 (w) = 1 or equivalently that

f 0 (w) =1

cos f (w).

38 BRUCE K. DRIVER

Now cos f (w) ∈ ¡1− w2¢1/2

, the question is which branch do we take. Todetermine this let us differentiate Eq. (13.2). Here we have

f 0 (w) =−i

iw +Q (1− w2)

½i+

1

2Q (1− w2)(−2w)

¾=

1

iw +Q (1− w2)

½1 + i

w

Q (1− w2)

¾=

1

iw +Q (1− w2)

(iw +Q

¡1− w2

¢Q (1− w2)

)=

1

Q (1− w2)

so we must use the same branch of the square-root used in Eq. (13.2).Hence we have shown

“d

dzsin−1 (z) =

1√1− z2

, ”

where one has to be careful about the branches which are used.

13.1. Complex and Contour integrals:

Definition 13.1. A path or contour C in D ⊂ C is a piecewise C1 — functionz : [a, b]→ C. For a function f : D→ C, we letZ

C

f (z) dz =

Z b

a

f (z (t)) z (t) dt

Example 13.2 (Some Contours). (1) z(t) = z0 + reit for 0 ≤ t ≤ π is a semi-circle centered at z0.

(2) If z0, z1 ∈ C then z (t) = z0 (1− t) + z1t for 0 ≤ t ≤ 1 parametrizes thestraight line segment going from z0 to z1.

(3) If z (t) = t+it2 for −1 ≤ t ≤ 1, then z (t) parametrizes part of the parabolay = x2. More generally z (t) = t+ if (t) parametrizes the graph, y = f (x) .

(4) z (t) = t+ i√1− t2 for −1 ≤ t ≤ 1 parametrizes the semicircle of radius 1

centered at 0 as does z (t) = e−iπt for −1 ≤ t ≤ 0.Example 13.3. Integrate f (z) = z − 1 along the two contours

(1) C1 : z = x for x = 0 to x = 2 and(2) C2 : z = 1 + eiθ for π ≤ θ ≤ 2π.For the first case we haveZ

C1

(z − 1) dz =Z 2

0

(x− 1) dx = 1

2(x− 1)2 |20 = 0

and for the second ZC2

(z − 1) dz =Z 2π

π

¡1 + eiθ − 1¢ ieiθdθ

=

Z 2π

π

iei2θdθ =i

2ei2θ|2ππ = 0


Example 13.4. Repeat the above example for f (z) = z − 1.For the first case we haveZ

C1

(z − 1) dz =Z 2

0

(x− 1) dx = 1

2(x− 1)2 |20 = 0

and for the second ZC2

(z − 1) dz =Z 2π

π

¡1 + e−iθ − 1¢ ieiθdθ

=

Z 2π

π

idθ = iπ 6= 0

Example 13.5 (I skipped this example.). Here we consider f (z) = y − x − i3x2

along the contours(1) C1 : consists of the straight line paths from 0→ i and i→ 1 + i and(2) C2 : consists of the straight line path from 0→ 1 + i.1. For the first case z = iy, dz = idy and z = x+ i and dz = dx, soZ

C1

f (z) dz =

Z 1

0

¡y − x− i3x2

¢ |x=0idy + Z 1

0

¡y − x− i3x2

¢ |y=1dx= i

Z 1

0

ydy +

Z 1

0

¡1− x− i3x2

¢dx

=i

2+

µ1− 1

2− i

¶=1

2(1− i) .

2. For the second contour, z = t (1 + i) = t+ it, then dz = (1 + i) dt,ZC2

f (z) dz =

Z 1

0

¡y − x− i3x2

¢ |x=y=t(1 + i)dt

= (1 + i)

Z 1

0

¡t− t− i3t2

¢dt

= (1 + i) (−i) = 1− i.

Notice the answers are different.

Example 13.6. Now lets use the same contours but with the function, f (z) = z2

instead. In this caseZC1

z2dz =

Z 1

0

(iy)2idy +

Z 1

0

(x+ i)2dx

= −i13+1

3(x+ i)

3 |10 = −i1

3+1

3(1 + i)

3 − 13i3

=1

3(1 + i)3

while for the second contour,ZC2

z2dz =

Z 1

0

t2 (1 + i)2 (1 + i) dt =1

3(1 + i)3 .

40 BRUCE K. DRIVER

14. (11/3/2003) Lecture 14 (Contour Integrals Continued)

Proposition 14.1. Let us recall some properties of complex integrals(1) Z b

a

w (φ (t)) φ (t) dt =

Z φ(b)

φ(a)

w (τ) dτ

(2) If f (z) is continuous in a neighborhood of a contour C, thenRCf (z) dz is

independent of how C is parametrized as long as the orientation is kept thesame.

(3) If −C denotes C traversed in the opposite direction, thenZ−C

f (z) dz = −ZC

f (z) dz.

Proof.(1) The first fact follows from the change of variable theorem for real variables.(2) Suppose that z : [a, b] → C is a parametrization of C, then other parame-

trizations of C are of the form

w(s) = z (φ (s))

where φ : [α, β]→ [a, b] such that φ (α) = a and φ (β) = b. HenceZ β

α

f (w (s))w0 (s) ds =Z β

α

f (z (φ (s))) z (φ (s))φ0 (s) ds

and letting t = φ (s) , we findZ β

α

f (w (s))w0 (s) ds =Z b

a

f (z (t)) z (t) dt

as desired.(3) Suppose that z : [0, 1]→ C is a parametrization of C, then w (s) := z (1− s)

parametrizes −C, so thatZ−C

f (z) dz = −Z 1

0

f (z (1− s)) z (1− s) ds =

Z 0

1

f (z (t)) z (t) dt

= −Z 1

0

f (z (t)) z (t) dt = −ZC

f (z) dz,

wherein we made the change of variables, t = 1− s.

Theorem 14.2 (Fundamental Theorem of Calculus). Suppose C is a contour inD and f : D→ C is an analytic function, thenZ

C

f 0 (z) dz = f(Cend)− f (Cbegin) .

Proof. Let z : [a, b]→ C parametrize the contour, thenZC

f 0 (z) dz =Z b

a

f 0 (z (t)) z (t) dt =Z b

a

d

dtf (z (t)) dt = f (z (t)) |t=bt=a.


Example 14.3. Using either of contours in Example 13.5, we again learn (moreeasily) Z

C1

z2dz =1

3z3|∂C1 =

1

3

h(1 + i)3 − (0)3

i.

Example 14.4. Suppose that C is a closed contour in C such which does not passthrough 0, then Z

C

zndz = 0 if n 6= −1.

The case n = 1 is different and leads to the winding number. This can becomputed explicitly, using a branch of a logarithm. For example if C : [0, 2π] →C\ 0 crosses (−∞, 0) only at z(0) = z(2π), thenZ

C

1

zdz = lim

ε↓0

ZCε

1

zdz = lim

ε↓0[Log(z (2π − ε))− Log(z (ε))]

= limε↓0

·ln

¯z (2π − ε)

z (ε)

¯+ i (2π −O (ε)−O (ε))

¸= i2π.

Also work out explicitly the special case where C (θ) = reiθ with θ : 0→ 2π.

Proposition 14.5. Let us recall some estimates of complex integrals

(1) ¯¯Z β

α

w (t) dt

¯¯ ≤

Z β

α

|w (t)| dt.

(2) We also have ¯ZC

f (z) dz

¯≤ZC

|f (z)| |dz| ≤ML

where |dz| = |z (t)| dt and M = supz∈C |f (z)| .Proof.

(1) To prove this let ρ ≥ 0 and θ ∈ R be chosen so thatZ β

α

w (t) dt = ρeiθ,

then¯¯Z β

α

w (t) dt

¯¯ = ρ = e−iθ

Z β

α

w (t) dt =

Z β

α

e−iθw (t) dt

=

Z β

α

Re£e−iθw (t)

¤dt ≤

Z β

α

¯Re£e−iθw (t)

¤¯dt

≤Z β

α

¯e−iθw (t)

¯dt =

Z β

α

|w (t)| dt.

42 BRUCE K. DRIVER

Alternatively:¯¯Z β

α

w (t) dt

¯¯ =

¯lim

mesh→0

Xw (ci) (ti − ti−1)

¯= limmesh→0

¯Xw (ci) (ti − ti−1)

¯≤ limmesh→0

X|w (ci)| (ti − ti−1) (by the triangle inequality)

=

Z β

α

|w (t)| dt.(2) For the last item¯Z

C

f (z) dz

¯=

¯¯Z b

a

f (z (t)) z (t) dt

¯¯ ≤

Z b

a

|f (z (t))| |z (t)| dt

≤M

Z b

a

|z (t)| dt ≤ML

wherein we have used

|z (t)| dt =q[x (t)]2 + [y (t)]2dt = d .


15. (11/05/2003) Lecture 15

Example 15.1. The goal here is to estimate the integral¯ZC

1

z4dz

¯where C is the contour joining i to 1 by a straight line path. In this case M =

1

| 12 (1+i)|4 and L = |1− i| = √2 and this gives the estimate¯ZC

1

z4dz

¯≤√2

1¯1√2

¯4 = 4√2.Example 15.2. Let C be the contour consisting of straight line paths −4 → 0,0→ 3i and then 3i→ −4 and we wish to estimate the integralZ

C

(ez − z) dz.

To do this notice that on C we have

|ez − z| ≤ |ez|+ |z| ≤ eRe z + |z| ≤ e0 + 4 = 5

while

(C) = 4 + 3 + |3i− (−4)| = 4 + 3 +p32 + 42 = 3 + 4 + 5 = 12

and hence ¯ZC

(ez − z) dz

¯≤ 12 · 5 = 60.

Note: The material after this point will not be on the second midterm.

Notation 15.3. Let D ⊂o C and α : [a, b]→ D and β : [a, b]→ D be two piecewiseC1 — contours in D. Further assume that either α (a) = β (a) and α (b) = β (b) orα and β are loops. We say α is homotopic to β if there is a continuos mapσ : [a, b]× [0, 1]→ D, such that σ (t, 0) = α (t) , σ (t, 1) = β (t) and either σ (a, s) =α (a) = β (a) and σ (b, s) = α (b) = β (b) for all s or t → σ (t, s) is a loop for all s.Draw lots of pictures here.

Definition 15.4 (Simply Connected). A connected region D ⊂o C is simplyconnected if all closed contours, C ⊂ D are homotopic to a constant path.

Theorem 15.5 (Cauchy Goursat Theorem). Suppose that f : D→ C is an analyticfunction and α and β are two contours in D which are homotopic relative end-pointsor homotopic loops in D, thenZ

α

f (z) dz =

Zβ

f (z) dz.

In particular if D is simply connected, thenZC

f (z) dz = 0

for all closed contours in D and complex analytic functions, f, on D.

Example 15.6. Suppose C is a closed contour in C, then(1)

RCesin zdz = 0 and

Rαesin zdz depends only on the endpoints of α.

(2)RCzndz = 0 for all n ∈ N∪ 0 .

44 BRUCE K. DRIVER

(3) However if C (θ) = reiθ for θ : 0→ 2π, thenZC

zdz =

Z 2π

0

re−iθireiθdθ = 2πir2 6= 0.

Example 15.7. Suppose C is a closed contour in C\ 0 , then(1)

RCesin zdz = 0 and

Rαesin zdz depends only on the endpoints of α.

(2) HoweverRCz−1dz = 2πi 6= 0.

(3) On the other hand if C is a loop in C \ (−∞, 0], then we knowZC

z−1dz = 0

this can be checked by direct computation. However it is harder to checkdirectly that Z

C

esin z

zdz = 0

for all closed contours in C \ (−∞, 0].


16. (11/07/2003) Lecture 16

Example 16.1 (Fourier Transform). So the exampleHC1zdz = 2πi again, where

C is a simple closed counter clockwise oriented contour surrounding 0. Do this bydeforming C to the unit circle contour.

Example 16.2 (Fourier Transform). The goal here is to compute the integral

Z :=

Z ∞−∞

e−12x

2

eiλxdx := limR→∞

Z R

−Re−

12x

2

eiλxdx.

We do this by completing the squares,

−12x2 + iλx = −1

2(x− iλ)2 − 1

2λ2

from which we learn

Z = e−12λ

2

limR→∞

Z R

−Re−

12 (x−iλ)2dx = e−

12λ

2

limR→∞

ZΓR

e−12z

2

dz

where ΓR (x) := x− iλ for x : −R→ R.We would like to replace the contour ΓR by[−R,R]. This can be done using the Cauchy Goursat theorem and the estimates,¯

¯Z ±R−iλ

±Re−

12 z

2

dz

¯¯ ≤ |λ| e− 1

2R2 → 0 as R→∞,

see Figure 3. Therefore we conclude that

Figure 3. By the Cauchy Goursat theorem, the integral of anyentire function around the closed countour shown is 0.

Z ∞−∞

e−12x

2

eiλxdx = e−12λ

2

limR→∞

Z R

−Re−

12x

2

dx =√2πe−

12λ

2

,

where the last integral is done by a standard real variable trick and the answer isgiven by

√2π.

Theorem 16.3 (Cauchy Integral Formula). Suppose that f : D → C is analyticand C ⊂ D is a contour which is homotopic to ∂D (z0, ε) in D \ z0 , then

(16.1)ZC

f (z)

z − z0dz = 2πif (z0) .

46 BRUCE K. DRIVER

Proof. Since z → f(z)z−z0 is analytic in D \ z0, the Cauchy Goursat Theorem

impliesZC

f (z)

z − z0dz =

Z∂D(z0,ε)

f (z)

z − z0dz =

Z 2π

0

f¡z0 + εeiθ

¢z0 + εeiθ − z0

iεeiθdθ

= i

Z 2π

0

f¡z0 + εeiθ

¢dθ → i

Z 2π

0

f (z0) dθ = 2πif (z0) as ε ↓ 0.

Example 16.4. Use complex methods to compute the integralZ ∞−∞

1

1 + x2dx = π.

This is done by the usual method, namely let CR (θ) = R eiθ for θ : 0→ π, then

limR→∞

¯ZCR

1

1 + z2kdz

¯≤ lim

R→∞1

1−R2kπR = 0

and therefore if we let ΓR be the contour [−R,R] followed by CR, we haveZ R

−R

1

1 + x2kdx = lim

R→∞

Z[−R,R]

1

1 + z2kdz = lim

R→∞

ZΓR

1

1 + z2kdz.

The last integral is independent of R > 1 and can be computed by deforming thecontours. For the first case we haveZ

ΓR

1

1 + z2=

ZΓR

1

(z − i) (z + i)dz = 2πi

1

i+ i= π.


17. (11/12/2003) Lecture 17

Gave the second midterm on Monday, 11/10/03.

Definition 17.1 (Residue). Suppose that D is a disk centered at z0 ∈ C andf : D \ z0→ C is analytic. The residue of f at z0 is defined by

resz0f :=1

2πi

I|z−z0|=ε

f (z) dz

where ε > 0 such that the contour |z − z0| = ε is in D. As usual the contour isgiven the counter clockwise orientation.

Lemma 17.2. Suppose that f is analytic inside a closed contour C and only 0 atone point z0 inside C and that f (z) = h(z)

g(z) where h and g are analytic functionsnear z0 and g0 (z0) 6= 0, then

resz0f :=h (z0)

g0 (z0).

Proof. Using results that we will prove shortly when considering power series,g (z) = (z − z0) k (z) where k is analytic near z0 and k (z0) = g0 (z0) 6= 0. Alterna-tively, use the fundamental theorem of calculus to write

g (z) = g (z)− g (z0) =

Z 1

0

d

dtg (z0(1− t) + tz) dt

=

Z 1

0

(z − z0) g0 (z0(1− t) + tz) dt

= (z − z0) k (z)

where k (z) :=R 10g0 (z0(1− t) + tz) dt. Then k (z0) = g0 (z0) and k is analytic with

k0 (z) =Z 1

0

g00 (z0(1− t) + tz) tdt

as we will show shortly below.Therefore by the Cauchy integral formula

1

2πi

I|z−z0|=ε

f (z) dz =1

2πi

I|z−z0|=ε

h (z)

g (z)dz

=1

2πi

I|z−z0|=ε

h (z) /k (z)

(z − z0)dz =

h (z0)

k (z0).

Since g0 (z0) = k (z0) , the result is proved.

Theorem 17.3 (Residue Theorem). Suppose that f : D \ z1, . . . , zn → C is ananalytic function and C is a simple counter clockwise closed contour in D such thatC “surrounds” z1, . . . , zn , thenZ

C

f (z) dz = 2πinXi=1

reszif.

Proof. The proof of this theorem is contained in Figure 4.

48 BRUCE K. DRIVER

Figure 4. Deforming a contour to circles around the singularitiesof f. The integration over the parts of the countour indicated bystraight lines cancel and so may be ignored.

Example 17.4. Use complex methods to compute the integral:Z ∞−∞

1

1 + x4dx =

π√2.

We will continue the method in Example 16.4 where it was already shown thatZ ∞−∞

1

1 + x4dx =

ZΓR

1

1 + z4dz

= 2πiX

z:z4+1=0 with z inside ΓR

resz1

1 + z4.

The roots of z4+1 = 0 inside of ΓR are z0 := eiπ/4 = 1√2(1 + i) and z1 := ei3π/4 =

1√2(−1 + i) , see Figure 5. When z is a root we have,

resz1

1 + z4=

1

4z3.

Hence Z ∞−∞

1

1 + x4dx = 2πi

·1

4z30+

1

4z31

¸=2πi

4

ne−i3π/4 + e−iπ/4

o=

πi

2√2−1− i+ 1− i

=π√2.

Example 17.5. Compute the integralZ ∞−∞

eiλx

1 + x2dx for λ ≥ 0.


Figure 5. Deforming contours to evaluted real integrals.

Close the contour in the upper half plane (note that¯eiλz

¯= eRe(iλz) = e−λy ≤ 1

for y ≥ 0) doing the usual estimates to show this OK. Hence we haveZ ∞−∞

eiλx

1 + x2dx =

ZΓR+[−R,R]

eiλz

1 + z2dz = 2πiresz=i

·eiλz

1 + z2

¸= 2πi

·eiλi

2i

¸= πe−λ.

Conclude from this that

πe−λ =Z ∞−∞

eiλx

1 + x2dx =

Z ∞−∞

cos (λx)

1 + x2dx+ i

Z ∞−∞

sin (λx)

1 + x2dx

and hence Z ∞−∞

cos (λx)

1 + x2dx = πe−λ andZ ∞

−∞

sin (λx)

1 + x2dx = 0.

50 BRUCE K. DRIVER

18. (11/14/2003) Lecture 18

Notation 18.1. We will write Ω ⊂o C if Ω is an open subset of C and f ∈ H(Ω)if f is analytic on Ω. Also let

D(z0, ρ) = z ∈ C : |z − z0| < ρD(z0, ρ) = z ∈ C : |z − z0| ≤ ρ and

∂D(z0, ρ) = z ∈ C : |z − z0| = ρ .18.1. On the proof of the Cauchy Goursat Theorem.

Theorem 18.2 (Differentiating under the integral sign). Suppose that f (t, z) is acontinuous function in (t, z) for a ≤ t ≤ b and z near z0 ∈ F where F = R or C.Further assume that ∂f(t,z)

∂z exists and is continuous in (t, z) with z near z0, then

d

dz

Z b

a

f (t, z) dt =

Z b

a

∂f (t, z)

∂zdt.

Moreover the function z → R ba∂f(t,z)∂z dt is continuous in z near z0.

Proof. (Sketch Briefly!) Let

F (z) :=

Z b

a

f (t, z) dt,

then(18.1)

F (z + h)− F (z)

h−Z b

a

∂f (t, z)

∂zdt =

Z b

a

·f (t, z + h)− f (t, z)

h− ∂f (t, z)

∂z

¸dt

andf (t, z + h)− f (t, z)

h=1

h

Z 1

0

d

dsf (t, z + sh) ds =

Z 1

0

∂f

∂z(t, z + sh) ds.

Therefore,¯f (t, z + h)− f (t, z)

h− ∂f (t, z)

∂z

¯=

¯Z 1

0

·∂f

∂z(t, z + sh)− ∂f (t, z)

∂z

¸ds

¯≤Z 1

0

¯∂f

∂z(t, z + sh)− ∂f (t, z)

∂z

¯ds

≤ maxs∈[0,1]

¯∂f

∂z(t, z + sh)− ∂f (t, z)

∂z

¯and the latter term goes to 0 uniformly in t as h → 0 by uniform continuity of∂f∂z (t, z) . Therefore we can let h→ 0 in Eq. (18.1) to findZ b

a

·f (t, z + h)− f (t, z)

h− ∂f (t, z)

∂z

¸dt→ 0 as h→ 0

and hence

limh→0

F (z + h)− F (z)

h=

Z b

a

∂f (t, z)

∂zdt.

The continuity in z is proved similarly,¯¯Z b

a

∂f (t, z + h)

∂zdt−

Z b

a

∂f (t, z)

∂zdt

¯¯ ≤

Z b

a

¯∂f (t, z + h)

∂z− ∂f (t, z)

∂z

¯dt→ 0 as h→ 0


by the same uniform continuity arguments used above.

Notation 18.3. Given D ⊂o C and a C2 map σ : [a, b] × [0, 1] → D, let σs :=σ(·, s) ∈ C1([a, b]→ D). In this way, the map σ may be viewed as a map

s ∈ [0, 1]→ σs := σ(·, s) ∈ C2([a, b]→ D),

i.e. s→ σs is a path of contours in D.

Definition 18.4. Given a regionD and α, β ∈ C2 ([a, b]→ D) , we will write α ' βin D provided there exists a C2 — map σ : [a, b] × [0, 1] → D such that σ0 = α,σ1 = β, and σ satisfies either of the following two conditions:

(1) ddsσ(a, s) =

ddsσ(b, s) = 0 for all s ∈ [0, 1], i.e. the end points of the paths

σs for s ∈ [0, 1] are fixed.(2) σ(a, s) = σ(b, s) for all s ∈ [0, 1], i.e. σs is a loop in D for all s ∈ [0, 1].See Figure 6.

Figure 6. Smooth homotopy of open paths and loops.

Proposition 18.5 (Baby Cauchy — Goursat Theorem). Let D be a region andα, β ∈ C2([a, b],D) be two contours such that α ' β in D. ThenZ

α

f(z)dz =

Zβ

f(z)dz for all f ∈ H(D) ∩ C1 (D) .

Proof. Let σ : [a, b]× [0, 1]→ D be as in Definition 18.4, then it suffices to showthe function

F (s) :=

Zσs

f(z)dz

is constant for s ∈ [0, 1]. For this we compute:

F 0(s) =d

ds

Z b

a

f(σ(t, s))σ(t, s)dt =

Z b

a

d

ds[f(σ(t, s))σ(t, s)] dt

=

Z b

a

f 0(σ(t, s))σ0(t, s)σ(t, s) + f(σ(t, s))σ0(t, s) dt

=

Z b

a

d

dt[f(σ(t, s))σ0(t, s)] dt

= [f(σ(t, s))σ0(t, s)]¯t=bt=a

= 0

52 BRUCE K. DRIVER

where the last equality is a consequence of either of the two endpoint assumptionsof Definition 18.4.Recall the Cauchy integral formula states,

(18.2) f (z) =1

2πi

ZC

f (w)

w − zdw

where C is simple closed contour in Ω traversed in the counter clockwise direction,z is inside C and f ∈ H (Ω) . Using the results in Proposition 18.5 we can rigorouslyprove Eq. (18.2) for f ∈ H (D) ∩ C1 (D) and for those contours C which are C2 —homotopic to ∂D (z, δ) for some δ > 0. Using Theorem 18.2, we may differentiateEq. (18.2) with respect to z repeatedly to learn the following theorem.

Theorem 18.6. Suppose f ∈ H (Ω) ∩ C1 (Ω) ,1 then f (n) exists and f (n) ∈ H (Ω)for all n ∈ N. Moreover if D is a disk such that D ⊂ Ω, then(18.3) f (n)(z) =

n!

2πi

I∂D

f(w)

(w − z)n+1dw for all z ∈ D.

Corollary 18.7 (Cauchy Estimates). Suppose that f ∈ H(Ω) where Ω ⊂o C andsuppose that D(z0, ρ) ⊂ Ω, then(18.4)

¯f (n)(z0)

¯≤ n!

ρnMρ

whereMρ := sup

|w−z0|=ρ|f(w)|

Proof. From Eq. (18.3) evaluated at z = z0 with C = ∂D(z0, ρ), we have¯f (n)(z0)

¯=

¯¯ n!2πi

IC

f(w)

(w − z0)n+1 dw

¯¯ ≤ n!

2π

Mρ

ρn+12πρ.

1As we will see later in Theorem 19.5, the assumption that f is C1 in this condition is redun-dant. Complex differentiability of f at all points z ∈ Ω already implies that f is C∞(Ω,C)!!


19. (11/17/2003) Lecture 19

Theorem 19.1 (Morera’s Theorem). Suppose that Ω ⊂o C and f ∈ C(Ω) is acomplex function such that

(19.1)Z∂T

f(z)dz = 0 for all solid triangles T ⊂ Ω,

then f ∈ H(Ω) and f (n) exists for all n, so f (n) ∈ H (Ω) for all n ∈ N ∪ 0 .Proof. Let D = D(z0, ρ) be a disk such that D ⊂ Ω and for z ∈ D let

F (z) =

Z[z0,z]

f(ξ)dξ

where [z0, z] is by definition the contour, σ(t) = (1 − t)z0 + tz for 0 ≤ t ≤ 1, seeFigure 7. For z ∈ D and h small so that z + h ∈ D we have, using Eq. (19.1),

ρ

Ω

Figure 7. Constructing a locally defined anti-derivative fo f toshow that f is analytic.

F (z + h)− F (z) =

Z[z,z+h]

f(w)dw =

Z 1

0

f(z + th)hdt = h

Z 1

0

f(z + th)dt

wherein we have parametrized [z, z + h] as w = z+ th. From this equation and thecontinuity of f,

F (z + h)− F (z)

h=

Z 1

0

f(z + th)dt→ f(z) as h→ 0.

Hence F 0 = f so that F ∈ H(D) ∩ C1 (D) . Theorem 18.6 now implies that F (n)

exists for all n and hence f (n) = F (n+1) ∈ H(D) exists for all n. Since D wasan arbitrary disk contained in Ω and the condition for being in H(Ω) is local weconclude that f ∈ H(Ω) and f (n) ∈ H (Ω) for all n.

54 BRUCE K. DRIVER

19.1. The material in this section was not covered in class.

Theorem 19.2 (A variant of Morera’s Theorem). (This may be skipped.) Supposethat f is a continuous function on a domain D such that

Rαf (z) dz only depends

on the end points of α, for example if D is simply connected and f is analytic onD. Then f has an anti-derivative, F. Namely, fix a z0 ∈ D and let Cz denote acontour in D such that Cz (0) = z0 and Cz (1) = z, then we may define

F (z) :=

ZCz

f (w) dw.

Proof. Let [z, z + h] denote the contour C (t) := z + th for t : 0→ 1. Then

F (z + h)− F (z) =

ZCz+h

f (w) dw −ZCz

f (w) dw

=

ZCz+h−Cz

f (w) dw =

Z[z,z+h]

f (w) dw

wherein the last equality we have used the fact that Cz+h − Cz and [z, z + h] arecontours with the same endpoints. Using this formula

F (z + h)− F (z)

h=1

h

Z 1

0

f (z + th)hdt

=

Z 1

0

f (z + th) dt→ f (z) as h→ 0.

The next theorem is the deepest theorem of this section.

Theorem 19.3 (Converse of Morera’s Theorem). Let Ω ⊂o C and f : Ω → C isa function which is complex differentiable at each point z ∈ Ω. Then H

∂T

f(z)dz = 0

for all solid triangles T ⊂ Ω.Proof. Write T = S1 ∪ S2 ∪ S3 ∪ S4 as in Figure 8 below.

Figure 8. Splitting T into four similar triangles of equal size.


Let T1 ∈ S1, S2, S3, S4 such that |R∂T

f(z)dz| = max| R∂Si

f(z)dz| : i =

1, 2, 3, 4, then

|Z∂T

f(z)dz| = |4X

i=1

Z∂Si

f(z)dz| ≤4Xi=1

|Z∂Si

f(z)dz| ≤ 4|Z∂T1

f(z)dz|.

Repeating the above argument with T replaced by T1 again and again, we find byinduction there are triangles Ti∞i=1 such that

(1) T ⊇ T1 ⊇ T2 ⊇ T3 ⊇ . . .(2) (∂Tn) = 2

−n (∂T ) where (∂T ) denotes the length of the boundary of T,(3) diam(Tn) = 2−n diam(T ) and

(19.2) |Z∂T

f(z)dz| ≤ 4n|Z∂Tn

f(z)dz|.

By finite intersection property of compact sets there exists z0 ∈∞Tn=1

Tn. Because

f(z) = f(z0) + f 0(z0)(z − z0) + o(z − z0)

we find¯¯4n Z

∂Tn

f(z)dz

¯¯ = 4n

¯¯ Z∂Tn

f(z0)dz +

Z∂Tn

f 0(z0)(z − z0)dz +

Z∂Tn

o(z − z0)dz

¯¯

= 4n

¯¯ Z∂Tn

o(z − z0)dz

¯¯ ≤ C n4

n

Z∂Tn

|z − z0| d |z|

where n → 0 as n→∞. SinceZ∂Tn

|z − z0| d |z| ≤ diam(Tn) (∂Tn) = 2−ndiam(T )2−n (∂T )

= 4−ndiam(T ) (∂T )

we see

4n

¯¯ Z∂Tn

f(z)dz

¯¯ ≤ C n4

n4−ndiam(T ) (∂T ) = C n → 0 as n→∞.

Hence by Eq. (19.2),R∂T

f(z)dz = 0.

The method of the proof above also gives the following corollary.

Corollary 19.4. Suppose that Ω ⊂o C is convex open set. Then for every f ∈ H(Ω)there exists F ∈ H(Ω) such that F 0 = f. In fact fixing a point z0 ∈ Ω, we may defineF by

F (z) =

Z[z0,z]

f(ξ)dξ for all z ∈ Ω.

By combining Theorem 19.1 and Theorem 19.3 we arrive at the important result.

56 BRUCE K. DRIVER

Theorem 19.5. Suppose that f ∈ H (Ω) , then f 0 ∈ H (Ω) and hence by induction,f (n) exists and f (n) ∈ H (Ω) for all n ∈ N∪ 0 .Exercise 19.6. Let Ω ⊂o C and fn ⊂ H(Ω) be a sequence of functions suchthat f(z) = limn→∞ fn(z) exists for all z ∈ Ω and the convergence is uniform oncompact subsets of Ω. Show f ∈ H(Ω) and f 0(z) = limn→∞ f 0n(z).Hint: Use Morera’s theorem to show f ∈ H(Ω) and then use Eq. (26.3) with

n = 1 to prove f 0(z) = limn→∞ f 0n(z).

19.2. More Applications of the Cauchy Goursat and the Cauchy IntegralFormula. The next two results were covered in class.

Corollary 19.7 ( Liouville’s Theorem). If f ∈ H(C) and f is bounded then f isconstant.

Proof. This follows from Eq. (18.4) with n = 1 and the letting n→∞ to findf 0(z0) = 0 for all z0 ∈ C.Corollary 19.8 (Fundamental theorem of algebra). Every polynomial p(z) of de-gree larger than 0 has a root in C.

Proof. Suppose that p(z) is polynomial with no roots in z. Then f(z) = 1/p(z)is a bounded holomorphic function and hence constant. This shows that p(z) is aconstant, i.e. p has degree zero.

19.2.1. The remainder of this subsection was not done in class.

Corollary 19.9 (Mean value property). Let Ω ⊂o C and f ∈ H(Ω), then f satisfiesthe mean value property

(19.3) f(z0) =1

2π

Z 2π

0

f(z0 + ρeiθ)dθ

which holds for all z0 and ρ ≥ 0 such that D(z0, ρ) ⊂ Ω.Proof. By Cauchy’s integral formula and parametrizing ∂D(z0, ρ) as z = z0 +

ρeiθ, we learn

f(z0) =1

2πi

Z∂D(z0,ρ)

f(z)

z − z0dz =

1

2πi

Z 2π

0

f(z0 + ρeiθ)

ρeiθiρeiθdθ

=1

2π

Z 2π

0

f(z0 + ρeiθ)dθ.

Proposition 19.10. Suppose that Ω is a connected open subset of C. If f ∈ H(Ω)is a function such that |f | has a local maximum at z0 ∈ Ω, then f is constant.

Proof. Let ρ > 0 such that D = D(z0, ρ) ⊂ Ω and |f(z)| ≤ |f(z0)| =: Mfor z ∈ D. By replacing f by eiθf with an appropriate θ ∈ R we may assumeM = f(z0). Letting u(z) = Re f(z) and v(z) = Im f(z), we learn from Eq. (19.3)that

M = f(z0) = Re f(z0) =1

2π

Z 2π

0

u(z0 + ρeiθ)dθ

≤ 1

2π

Z 2π

0

min(u(z0 + ρeiθ), 0)dθ ≤M


since¯u(z0 + ρeiθ)

¯ ≤ ¯f(z0 + ρeiθ)¯ ≤ M for all θ. From the previous equation it

follows that

0 =

Z 2π

0

©M −min(u(z0 + ρeiθ), 0)

ªdθ

which in turn implies that M = min(u(z0 + ρeiθ), 0), since θ → M − min(u(z0 +ρeiθ), 0) is positive and continuous. So we have proved M = u(z0 + ρeiθ) for all θ.Since

M2 ≥ ¯f(z0 + ρeiθ)¯2= u2(z0 + ρeiθ) + v2(z0 + ρeiθ) =M2 + v2(z0 + ρeiθ),

we find v(z0 + ρeiθ) = 0 for all θ. Thus we have shown f(z0 + ρeiθ) = M for all θand hence by Corollary 27.8, f(z) =M for all z ∈ Ω.The following lemma makes the same conclusion as Proposition 19.10 using the

Cauchy Riemann equations. This Lemma may be skipped.

Lemma 19.11. Suppose that f ∈ H(D) where D = D(z0, ρ) for some ρ > 0. If|f(z)| = k is constant on D then f is constant on D.

Proof. If k = 0 we are done, so assume that k > 0. By assumption

0 = ∂k2 = ∂ |f |2 = ∂(ff) = ∂f · f + f∂f

= f∂f = ff 0

wherein we have used

∂f =1

2(∂x − i∂y) f =

1

2(∂x + i∂y) f(z) = ∂f = 0

by the Cauchy Riemann equations. Hence f 0 = 0 and f is constant.

Corollary 19.12 (Maximum modulous principle). Let Ω be a bounded region andf ∈ C(Ω) ∩ H(Ω). Then for all z ∈ Ω, |f(z)| ≤ sup

z∈∂Ω|f(z)| . Furthermore if there

exists z0 ∈ Ω such that |f(z0)| = supz∈∂Ω

|f(z)| then f is constant.

Proof. If there exists z0 ∈ Ω such that |f(z0)| = maxz∈∂Ω |f(z)| , then Proposi-tion 19.10 implies that f is constant and hence |f(z0)| = sup

z∈∂Ω|f(z)| . If no such z0

exists then |f(z)| ≤ supz∈∂Ω

|f(z)| for all z ∈ Ω.

19.3. Series.

Definition 19.13. Given a sequence zn∞n=0 , we say the sum,P∞

n=0 zn =: Sexists or

P∞n=0 zn is convergent if the sequence

SN :=NXn=0

zn → S as N →∞.

We say thatP∞

n=0 zn is absolutely convergent ifP∞

n=0 |zn| <∞.

Remark 19.14. Since limN→∞ SN = S iff limN→∞ReSN = ReS and limN→∞ReSN =ReS, it follows that

P∞n=0 zn exists iff

P∞n=0Re zn and

P∞n=0 Im zn exists. In this

case ∞Xn=0

zn =∞Xn=0

Re zn +∞Xn=0

Im zn.

58 BRUCE K. DRIVER

Proposition 19.15 (Completeness of C). IfP∞

n=0 |zn| < ∞ thenP∞

n=0 zn existsand

(19.4)

¯¯∞Xn=0

zn

¯¯ ≤

∞Xn=0

|zn| .

Proof. (Skip the proof and just take this as a basic fact.) Because of Remark19.14 and the estimates,

∞Xn=0

|Re zn| ≤∞Xn=0

|zn| <∞ and∞Xn=0

|Im zn| ≤∞Xn=0

|zn| <∞

it suffices to consider the real case. Now for M > N we have

|SN − SM | =¯¯

NXn=M+1

zn

¯¯ ≤

NXn=M+1

|zn| ≤∞X

n=M+1

|zn|

=∞Xn=0

|zn|−MXn=0

|zn|→ 0 as M,N →∞.

Therefore by the basic “completeness” of the real numbers, limN→∞ SN exists. Forthe estimate in Eq. (19.4), we have¯

¯∞Xn=0

zn

¯¯ = lim

N→∞

¯¯NXn=0

zn

¯¯ ≤ lim

N→∞

NXn=0

|zn| =∞Xn=0

|zn| .

Example 19.16. Let z ∈ C and let us consider the geometric series P∞n=0 zn. Inthis case we may find the partial sums, SN :=

PNn=0 z

n explicitly since

SN − zSN = 1− zN+1.

Solving this equation for SN then implies

SN =

(1−zN+1

1−z if z 6= 1N + 1 if z = 1.

From this expression we see thatP∞

n=0 zn exists iff |z| < 1 and in which case

∞Xn=0

zn =1

1− z.

Let us note that

RN (z) =1

1− z−

NXn=0

zn =1

1− z− 1− zN+1

1− z

=zN+1

1− z

so that

(19.5)1

1− z=

NXn=0

zn +RN (z)


and

|RN (z)| = |z|N+1|1− z| ≤

|z|N+11− |z| for |z| < 1.

Remark 19.17. More generally the same argument shows

(19.6)mXk=n

zk =zm+1 − zn

z − 1 if z 6= 1.

Example 19.18. Also showed

1

1 + z2=∞Xn=0

(−1)n z2n for |z| < 1

and explained why the series only converges for |z| < 1 by looking at the locationof the poles of the function f (z) = 1

1+z2 .

60 BRUCE K. DRIVER

20. (11/19/2003) Lecture 20

Theorem 20.1 (Differentiating and integrating a sum of analytic functions). Sup-pose that fn : Ω→ C is a sequence of analytic functions such that

|fn (z)| ≤Mn for all n ∈ N and z ∈ Cwhere

P∞n=1Mn <∞. Then

(1) If C is any contour in Ω, we haveZC

F (z) dz =∞Xn=1

ZC

fn (z) dz.

(2) The function F (z) :=P∞

n=1 fn (z) is an analytic.(3) F 0 (z) =

P∞n=1 f

0n (z) and in fact

(20.1) F (k) (z) =∞Xn=1

f (k)n (z) for all k ∈ N0 and z ∈ Ω.

Part of the assertion here is that all sums appearing are absolutely convergent.

Proof. Later.

Remark 20.2 (Theorem 20.1 does not hold for real variable functions). It shouldbe noted that Eq. (20.1) is not correct when z is replace by a real variable. Forexample, the series

(20.2) F (x) :=∞Xn=1

sinnx

n2

is perfectly convergent for all x ∈ R, however if we differentiate it once or twice weget

∞Xn=1

cosnx

nand −

∞Xn=1

sinnx

which are no longer convergent. To understand a little better what is going on,notice that the series

∞Xn=1

sinnz

n2=∞Xn=1

einxe−ny + einxeny

2in2

is not convergent if z = x+ iy with y 6= 0. This is simply because,

limn→∞

¯einxe−ny + einxeny

2in2

¯=∞ in this case.

Example 20.3. Differentiating the formula,

1

1− z=∞Xn=0

zn for |z| < 1

gives1

(1− z)2=∞Xn=0

nzn−1

and differentiating again gives

2!

(1− z)3=∞Xn=0

n (n− 1) zn−2


and repeating to get

m!

(1− z)m+1 =

∞Xn=0

[n (n− 1) · · · (n−m+ 1)] zn−m.

Example 20.4. Integrating the formula

− d

dzLog (1− z) =

1

1− z=∞Xn=0

zn

implies

−Log (1− z) =

Z z

0

− d

dwLog (1− w) dw =

Z z

0

∞Xn=0

wndw

=∞Xn=0

(−1)nZ z

0

wndw

=∞Xn=0

1

n+ 1zn+1 if |z| < 1.

which implies

Log (1− z) = −∞Xn=0

1

n+ 1zn+1 if |z| < 1.

Theorem 20.5 (Taylor’s Theorem). Let Ω ⊂o C be an open set, f ∈ H(Ω) andD = D(z0, ρ) is a disk such that D ⊂ Ω then

(20.3) f(z) =∞Xn=0

an(z − z0)n for all z ∈ D.

where

(20.4) an =f (n)(z0)

n!=

1

2πi

I∂D

f(w)

(w − z0)n+1dw.

Proof. Let g (z) := f (z0 + z) and r < ρ and |z| < r, then for |w| = r,

1

w − z=1

w

1

1− z/w=1

w

∞Xn=0

³ zw

ń.

Applying Theorem 20.1 with fn (w) = g (w) 1w¡zw

¢nandMn = maxD |f | 1r

¡ρr

¢nwe

find using the Cauchy integral formula that

f (z0 + z) = g (z) =1

2πi

I|w|=r

g (w)

(w − z)dw =

1

2πi

I|w|=r

∞Xn=0

g (w)1

w

³ zw

ńdw

=∞Xn=0

"1

2πi

I|w|=r

g (w)

wn+1dw

#zn =

∞Xn=0

anzn(20.5)

where

an =1

2πi

I|w|=r

g (w)

wn+1dw =

g(n) (0)

n!=

f (n) (z0)

n!

=1

2πi

I|w−z0|=r

f (w)

(w − z0)n+1 dw.

62 BRUCE K. DRIVER

Replacing z by z − z0 in Eq. (20.5) completes the proof of Eq. (20.3).

Example 20.6. (1) Suppose f (z) = 11−z , then f (n) (z) = n!

(1−z)n+1 and hence

f (n) (0) = n! and we find again that

1

1− z=∞Xn=0

n!

n!zn =

∞Xn=0

zn.

(2) Suppose we wish to find the power series expansion of f (z) = 11−z centered

at 3 which will necessarily converge if |z − 3| < 2. To do this write z = 3+hso that

f (z) =1

1− (3 + h)=

1

−2− h= −1

2

1

1 + h/2= −1

2

∞Xn=0

µ−h2

¶n=∞Xn=0

(−12)n+1 (z − 3)n .

Moral: try to avoid computing derivatives whenever possible.(3) Since dn

dzn ez = ez for all n,

ez =∞Xn=0

1

n!zn

which is convergent for all z ∈ C since ez is entire.(4) By substituting −z for z,

e−z =∞Xn=0

(−1)nn!

zn.

and z3 for z we find

ez3

=∞Xn=0

1

n!z3n

(5) Since

ez0+h = ez0eh =∞Xn=0

ez0hn

n!

we have writing z = z0 + h, that

ez =∞Xn=0

ez0(z − z0)

n

n!.

(6) Similarly,

eiz =∞Xn=0

(iz)n

n!=∞Xn=0

(i)n

n!zn =

∞Xn=0

(i)2n

(2n)!z2n +

∞Xn=0

(i)2n+1

(2n+ 1)!z2n+1

=∞Xn=0

(−1)n(2n)!

z2n + i∞Xn=0

(−1)n(2n+ 1)!

z2n+1.

From this we deduce that

sin z =eiz − e−iz

2i=∞Xn=0

(−1)n(2n+ 1)!

z2n+1


and

cos z =eiz + e−iz

2=∞Xn=0

(−1)n(2n)!

z2n.

(7) Similarly,

sinh z =∞Xn=0

1

(2n+ 1)!z2n+1 and

cosh z =∞Xn=0

1

(2n)!z2n.

(8) Do sin (z) centered at z0. To do this again let z = z0 + h and then

sin (z) = sin (z0 + h) = sin (z0) sin (h) + cos (z0) cos (h)

= sin (z0)∞Xn=0

(−1)n(2n+ 1)!

(z − z0)2n+1

+ cos (z0)∞Xn=0

(−1)n(2n)!

(z − z0)2n

.

(9) Consider the power series expansion of the function f (z) := 1w−z centered

at z = z0 with z0 6= w. To do this again write z = z0 + h ∈ C, then1

w − z=

1

w − z0 − h=

1

w − z0

1

1− h/(w − z0)

=1

w − z0

∞Xn=0

µh

w − z0

¶n=∞Xn=0

µ1

w − z0

¶n+1(z − z0)

n

provided that |h| = |z − z0| < |w − z0| , i.e.

(20.6)1

w − z=∞Xn=0

µ1

w − z0

¶n+1(z − z0)

n for |z − z0| < |w − z0| .

We may now also differentiate this series in z to learn

1

(w − z)2 =

∞Xn=0

n

µ1

w − z0

¶n+1(z − z0)

n−1

and again to learn

2

(w − z)3=∞Xn=0

n (n− 1)µ

1

w − z0

¶n+1(z − z0)

n−2

64 BRUCE K. DRIVER

21. (11/21/2003) Lecture 21

• Reviewed the general method of using the residue theorem for computingreal integrals. So far we are restricted to computing residues only in simplecontexts. This will be remedied in the next lecture.

Example 21.1. Suppose that f (z) = (1 + z)α := eαLog(1+z). Then

f 0 (z) = α (1 + z)α−1 , f 00 (z) = α (α− 1) (1 + z)α−2 , . . .

f (n) (z) = α (α− 1) . . . (α− n+ 1) (1 + z)α−n

and therefore,

(1 + z)α =∞Xn=0

α (α− 1) . . . (α− n+ 1)

n!zn.

For example if α = −1, thenα (α− 1) . . . (α− n+ 1) = −1 (−2) . . . (−n) = (−1)n n!

and we find again that1

1 + z=∞Xn=0

(−1)n zn.

Suppose that f is an analytic function near z = z0 and f (z0) = 0. Then if f isnot identically zero, there is a first n ∈ N such that f (n) (z0) 6= 0, and therefore fhas a power series expansion of the form

f (z) =∞Xk=n

ak (z − z0)k= (z − z0)

n∞Xk=n

ak−nk (z − z0) = (z − z0)ng (z)

where g is analytic on the same set where f is analytic and g (z0) 6= 0. In this casewe say the f has a zero of order n at z0. Notice that if f has a zero of order∞ at z0 then f = 0 near z0 and in fact f = 0 on the connected component of Ωcontaining z0.

Theorem 21.2 (Analytic Continuation). Suppose that f : Ω → C is an analyticfunction on a connected open subset C such that Z (f) := z ∈ Ω : f (z) = 0 hasan accumulation point in Ω, then f ≡ 0.Proof. Suppose for simplicity 0 ∈ Ω and there exists zn ∈ Ω such that zn 6= 0

for all n and zn → 0 as n→∞. Writing

f (z) =∞Xn=0

anzn,

we have 0 = f (zn)→ f (0) = a0. Since

f (z)

z=∞Xn=1

anzn−1

we have 0 = f(zn)zn→ a1 showing a1 = 0. Hence now we have

f (z)

z2=∞Xn=2

anzn−2


and 0 = f(zn)z2n→ a2 showing a2 = 0. Continuing this way, we learn an = 0 for all

n and hence f (z) = 0 near 0. Since Ω is connected we may connected any pointz ∈ Ω by a path and then use the Picture in Figure 9 below to argue that f (z) = 0,i.e. f ≡ 0.

Figure 9. Stringing together a sequence of disks in order to showthat if f = 0 near one point in a connected region then f ≡ 0.

Corollary 21.3. If f and g are two entire functions such that f = g on the realaxis then f (z) = g (z) for all z ∈ C. In particular if f (x) is a real valued functionof x ∈ R, there is at most one extension of f to an analytic function on C.Proof. Apply the previous theorem to f − g.

Example 21.4. Suppose we wish to verify that

(21.1) sin (z + w) = sin z cosw + cos z sinw

using only the statement for real z and w. To do this, first assume that w is real,then both sides of Eq. (21.1) are analytic in z and agree for z real and thereforeare equal for all z ∈ C.Hence we now know that Eq. (21.1) holds for w ∈ R andz ∈ C. Now fix z ∈ C, then both sides of Eq. (21.1) are analytic in w and agree forw real and therefore are equal for all w ∈ C.

66 BRUCE K. DRIVER

22. (11/24/2003) Lecture 22

22.1. Laurent Series and Residues. For z0 ∈ C and 0 ≤ r < R ≤ ∞, let

A (z0, r, R) := z ∈ C : r < |z − z0| < Rso that A (z0, r, R) is an annulus centered at z0.

Theorem 22.1 (Laurent Series). Suppose that f : A (z0, r, R)→ C is analytic andlet

(22.1) an =1

2πi

ICρ

f (z)

(z − z0)n+1 dz.

where Cρ (θ) := z0 + ρeiθ for θ : 0→ 2π and r < ρ < R. Then

(22.2) f (z) =∞X

n=−∞an (z − z0)

n for all z ∈ A (z0, r, R)

where the above Laurent series converges absolutely.

Proof. (Only Sketched in Class.) First suppose that z0 = 0 and let z ∈A (0, r, R) . Choose r0 and R0 such that r < r0 < |z| < R0 < R. In this casethe Cauchy integral formula may be written as

(22.3) f (z) =1

2πi

"ICR0

f (w)

(w − z)dw −

ICr0

f (w)

(w − z)dw

#,

see Figure 10.

Figure 10. Setting up to use the Cauchy integral formula.


For w ∈ CR0we have

1

w − z=1

w

1

1− z/w=1

w

∞Xn=0

³ zw

ńand hence applying Theorem 20.1 we find

1

2πi

ICR0

f (w)

(w − z)dw =

∞Xn=0

"1

2πi

I|w|=R0

f (w)

wn+1dw

#zn.

By Cauchy Goursat theorem,

1

2πi

I|w|=R0

f (w)

wn+1dw =

1

2πi

I|w|=ρ

f (w)

wn+1dw = an

and so

(22.4)1

2πi

ICR0

f (w)

(w − z)dw =

∞Xn=0

anzn.

Similarly for w ∈ Cr0 ,

− 1

w − z=1

z

1

1− w/z=1

z

∞Xn=0

³wz

ńand hence applying Theorem 20.1,

− 1

2πi

ICr0

f (w)

(w − z)dw =

∞Xn=0

"1

2πi

ICr0

f (w)wndw

#z−(n+1)

=∞Xn=0

"1

2πi

ICρ

f (w)

w−ndw

#z−(n+1)

where we have used the Cauchy Goursat theorem in the last equality. Finally lettingk = − (n+ 1) or −n = k + 1 in the previous sum gives

(22.5) − 1

2πi

ICr0

f (w)

(w − z)dw =

−1Xk=−∞

"1

2πi

ICρ

f (w)

wk+1dw

#zk =

−1Xk=−∞

akzk.

Combining Eqs. (22.3), (22.4) and (22.5) verifies Eq. (22.2) when z0 = 0.When z0 6= 0, apply what we have just proved to g (h) = f (z0 + h) with h =

(z − z0) to learn

f (z) = g (h) =∞X

n=−∞anh

n =∞X

n=−∞an (z − z0)

n

where

an :=1

2πi

I|w|=ρ

g (w)

wn+1dw =

1

2πi

I|w|=ρ

f (z0 + w)

wn+1dw.

Finally make the change of variables, z = z0 + w in the previous integral to learn

an =1

2πi

I|z−z0|=ρ

f (z)

(z − z0)n+1 dz.

68 BRUCE K. DRIVER

Definition 22.2. An analytic function f on Ω \ z0 is said to have an isolatedsingularity at z0. Let

f (z) =∞X

n=−∞an (z − z0)

n for 0 < |z − z0| < R

be the Laurent series expansion of f centered at z0. The portion of this sum:

−1Xn=−∞

an (z − z0)n

is called principle part of f near z0.(1) f has an essential singularity at z0 if # n < 0 : an 6= 0 =∞.(2) f has a removable singularity at z0 if # n < 0 : an 6= 0 = 0, i.e. if the

principle part is 0.(3) f has a pole of order N if the principle part of f at z0 is of the form

−1Xn=−N

an (z − z0)n with a−N 6= 0.

If N = 1, f is said to have a simple pole at z0.

Remark 22.3. If f has an isolated singularity at z0, then by Eq. (22.1)

resz0f = a−1.

Example 22.4.(1) e1/z =

P∞n=0

1n!z−n, res0

¡e1/z

¢= 1 and z = 0 is an essential singularity

point. Mention consequences for integrals.(2) Since

sin z

z3=1

z3

µz − z3

3!+

z5

5!− . . .

¶= z−2 − 1

3!+

z2

5!− . . . ,

res0£sin zz3

¤= 0 and there is a pole of order 2 at 0.

(3)

cos z

z3=1

z3

µ1− z2

2!+

z4

4!− . . .

¶= z−3 − z−1

2!+

z

4!− . . .

so that res0£cos zz3

¤= −12 .

(4) Use partial fractions to find the Laurent series expansions of

f (z) :=1

(z − 1) (z − 2) at z0 = 0

in the three regions: |z| < 1, 1 < |z| < 2 and |z| > 2.To do this write1

(z − 1) (z − 2) =A

z − 1 +B

z − 2


and multiply by (z − 1) (z − 2) to find1 = A (z − 2) +B (z − 1) .

Evaluating at z = 1 and z = 2 then implies that A = −1 and B = 1 so that1

(z − 1) (z − 2) =−1z − 1 +

1

z − 2 .

(a) For |z| < 1, we havef (z) =

1

1− z− 12

1

1− z/2

=∞Xn=0

·zn − 1

2

³z2

ń¸=∞Xn=0

·1− 1

2n

¸zn.

(b) For 1 < |z| < 2, we havef (z) = −1

z

1

1− 1/z −1

2

1

1− z/2

= −1z

∞Xn=0

z−n − 12

∞Xn=0

³z2

ń= −

∞Xn=0

z−n−1 − 12

∞Xn=0

³z2

ń.

(c) For |z| > 2, we havef (z) = −1

z

1

1− 1/z +1

z

1

1− 2/z

= −∞Xn=0

z−n−1 +1

z

∞Xn=0

µ2

z

¶n=∞Xn=0

(2n − 1) z−n−1.

70 BRUCE K. DRIVER

23. (11/26/2003) Lecture 23

Proposition 23.1. Suppose f (z) is analytic on D0 (z0, ε) and

(23.1) f (z) =NX

n=−Kan (z − z0)

n+O

³(z − z0)

N+1´,

then

an =1

2πi

I|z−z0|=ρ

f (z)

(z − z0)n+1 dz with 0 < ρ < ε

are the Laurent coefficients of f for n = −K, . . . ,N. In particular if f is analyticon D0 (z0, ε) and

(23.2) f (z) =NXn=0

ann (z − z0) +O³(z − z0)

N+1´,

then an =f(n)(z0)

n! for n ≤ N.

Proof. For simplicity of notation let us assume that z0 = 0. Using

1

2πi

I|z|=ρ

1

wndw =

½0 if n 6= 11 if n = 1,

1

2πi

I|z|=ρ

f (z)

zk+1dz =

1

2πi

I|z|=ρ

1

zk+1

ÃNX

n=−Kanz

n +O¡zN+1

¢!dz

=1

2πi

I|z|=ρ

1

zk+1

ÃNX

n=−Kanz

n +O¡zN+1

¢!dz

= ak +1

2πi

I|z|=ρ

O¡zN−k

¢dz

which gives the result, since¯¯ 12πi

I|z|=ρ

O¡zN−k

¢dz

¯¯ ≤

¯1

2πO¡ρN−k

¢¯2πρ→ 0 as ρ→ 0

provided N ≥ k. The second assertion follows from the first using

f (n) (0) =n!

2πi

I|z|=ρ

f (z)

zn+1dz for n ≥ 0.

Here is a second proof of the second assertion. We haveNXn=0

anzn +O

¡zN+1

¢= f (z) =

∞Xn=0

f (n) (0)

n!zn

and therefore,NXn=0

·f (n) (0)

n!− an

¸zn = O

¡zN+1

¢− zN+1∞X

n=N+1

f (n) (0)

n!zn−N−1 = O

¡zN+1

¢.

Taking z = 0 shows f(0)(0)0! − a0 = 0 and then working inductively we learn an =

f(n)(0)n! for all n ≤ N.


Remark 23.2. In working the following examples we will make use of the followingbasic power series:

(1) (1 + z)α=P∞

n=0α(α−1)...(α−n+1)

n! zn so that

(1 + z)α = 1 + αz +α (α− 1)

2!z2 +

α (α− 1) (α− 2)3!

z2 + . . . .

(2) ez =P∞

n=01n!z

n

(3) sin z =P∞

n=0(−1)n(2n+1)!z

2n+1

(4) cos z =P∞

n=0(−1)n(2n)! z

2n

(5) sinh z =P∞

n=01

(2n+1)!z2n+1

(6) cosh z =P∞

n=01

(2n)!z2n.

Example 23.3. #1 on p. 238. Find the order m and residue B of the poles of thefollowing functions

1)z2 + 2

z − 1 , 2)

µz

2z + 1

¶3, 3)

ez

z2 + π24)

ez

(z2 + 1)2

5)1

sin (z2), 6)

ez

sin2 (z)at z = π

(1) Let z = 1 + h, then

z2 + 2

z − 1 =(1 + h)2 + 2

h=3 + 2h+ h2

h

so m = 1 and B = 3. Alternatively, res1³z2+2z−1

´= 12+2

1 = 3.

(2) Let z = −12 + h, thenµz

2z + 1

¶3=

µ−12 + h

2h

¶3=

1

8h3

µ−12+ h

¶3=

1

8h3

µ−18+3

4h− 3

2h2 + h3

¶so that m = 3 and B = − 3

16 .(3) The poles is at ±iπ and is of order 1 and we have

res±iπe±iπ

2 (±iπ) = ±i

2π.

(4) The poles are at ±i and m = 2. Let z = i+ h to find

ez

(z2 + 1)2=

ez

(z + i)2 (z − i)2=

ei+h

(2i+ h)2h2=

ei

−41

h2eh(1 + h/2i)−2

=ei

−41

h2(1 + h+O

¡h2¢)(1− 2h/2i+O

¡h2¢)

=ei

−41

h2(1 + h+O

¡h2¢)(1 + ih+O

¡h2¢)

=

µ· · ·+ ei

−4 (i+ 1)h−1 + . . .

¶and so

resiez

(z2 + 1)2= −e

i

4(i+ 1) .

72 BRUCE K. DRIVER

(5) This function has poles at z2 = nπ or z = (nπ)1/2 . Let z0 ∈ (nπ)1/2 andz = z0 + h. Then

sin¡z2¢= sin

¡z20 + 2z0h+ h2

¢= sin

¡nπ + 2z0h+ h2

¢= sin (nπ) cos

¡2z0h+ h2

¢+ cos (nπ) sin

¡2z0h+ h2

¢= (−1)n ¡2z0h+O

¡h2¢¢

where we assume that n 6= 0 for the moment. Then1

sin (z2)=

1

(−1)n (2z0h+O (h2))=(−1)n2z0h

1

1 +O (h)

=(−1)n2z0h

(1 +O (h))

and so

res(nπ)1/21

sin (z2)=

(−1)n2 (nπ)1/2

if n 6= 0.This can be done by our old friend as well, namely,

res(nπ)1/21

sin (z2)=

1

2z cos (z2)|z=(nπ)1/2 =

(−1)n2 (nπ)1/2

.

For n = 0, we have again that

sin¡z2¢= z2 − z6

6+O

¡z10¢

so that1

sin (z2)=

1

z2 − z6

6 +O (z10)=1

z21

1− z4

6 +O (z8)=1

z2¡1 +O

¡z4¢¢

and sores0

1

sin (z2)= 0.


24. (12/1/2003) Lecture 24

Example 24.1. (Skipped.) The function ez

sin2(z)at z = π has a pole of order 2

since sin2 (z) has a zero of order two there. So again let z = π + h and use

sin (z) = sin (π + h) = − sin (h)so that

ez

sin2 (z)=

eπ+h

sin2 (π + h)= eπ

³1 + h+ h2

2! + . . .´

sin2 (h)

= eπ

³1 + h+ h2

2! + . . .´

(h− h3/3! + . . . )2=

eπ

h2

³1 + h+ h2

2! + . . .´

(1− h2/3! + . . . )2

=eπ

h2¡1 + h+O

¡h2¢¢ ¡

1 +O¡h2¢¢

and so

resπez

sin2 (z)= eπ.

Example 24.2. (Skipped.) Find the first few terms in the Taylor series expansionof 1

1−z cos z and11−z

1cos z . To do this we have

1

1− zcos z =

¡1 + z + z2 + z3 +O

¡z4¢¢µ

1− z2

2!+O

¡z4¢¶

= 1− z2

2!+ z − z3

2!+ z2 + z3 +O

¡z4¢= 1 + z +

z2

2+

z3

2+O

¡z4¢

and similarly,

1

1− z

1

cos z=¡1 + z + z2 + z3 +O

¡z4¢¢ 1¡

1− z2

2! +O (z4)¢

=¡1 + z + z2 + z3 +O

¡z4¢¢µ

1 +z2

2!+O

¡z4¢¶

= 1 +z2

2!+ z +

z3

2!+ z2 + z3 +O

¡z4¢

= 1 + z +3

2z2 +

3

2z3 +O

¡z4¢.

Example 24.3. (Skipped.) Now consider1

1− z

1

sin z=

1

1− z

1¡z − z3

3! +O (z5)¢

=1

z

1¡1− z2

3! +O (z4)¢ ¡1 + z + z2 + z3 +O

¡z4¢¢

=1

z

µ1 +

z2

3!+O

¡z4¢¶ ¡

1 + z + z2 + z3 +O¡z4¢¢

=1

z

µ1 + z + z2 + z3 +

z2

3!+

z3

3!+O

¡z4¢¶

=1

z+ 1 +

7

6

¡z + z2

¢+O

¡z3¢.

74 BRUCE K. DRIVER

Example 24.4. (Skipped.) Show

resz=0ez

z sin z= resz=0

ez

sin2 z= 0.

To do this we have

sin2 z =¡z − z3/3! +O

¡z5¢¢2

=¡z − z3/3! +O

¡z5¢¢ ¡

z − z3/3! +O¡z5¢¢

= z2 +O¡z4¢= z2

¡1 +O

¡z2¢¢

and therefore

1

sin2 z=

1

z2 (1 +O (z2))=1

z2¡1 +O

¡z2¢¢= z−2 +O (1) .

Thereforeez

sin2 z=¡z−2 +O (1)

¢ ¡1 +O

¡z2¢¢= z−2 +O (1)

Similarly

z sin z = z¡z − z3/3! +O

¡z5¢¢= z2

¡1 +O

¡z2¢¢

and so the residue is the same.

Example 24.5. ShowI|z|=1

ez

z sin2 zdz = 2πi · resz=0 ez

z sin3 z= 2πi

5

6=5

3πi.

To do this we have

z sin2 z =¡z − z3/3! +O

¡z5¢¢2

= z¡z − z3/3! +O

¡z5¢¢ ¡

z − z3/3! +O¡z5¢¢

= z¡z2 − 2z4/3! +O

¡z5¢¢

= z3 − 13z5 +O

¡z6¢= z3

µ1− 1

3z2 +O

¡z3¢¶

and hence

1

z sin2 z=1

z31¡

1− 13z2 +O (z3)

¢=1

z3

Ã1 +

1

3z2 +O

¡z3¢+

µ1

3z2 +O

¡z3¢¶2!

= z−3 +1

3z−1 +O (1) .

Hence

1

z sin2 zez =

·z−3 +

1

3z−1 +O (1)

¸ £1 + z + z2/2 +O

¡z3¢¤

= . . .

µ1

3+1

2

¶z−1 + . . .

and

resz=0ez

z sin3 z=5

6


Example 24.6. #4 p. 219, I|z|=1

1

z2 sinh zdz = −πi

3

To see this we need to compute the residue at 0. For this we have

1

z2 sinh z=

1

z2¡z + z3

3! +z5

5! + . . .¢ = 1

z31

1 + z2

3! +z5

5! + . . .

=1

z3

Ã1−

µz2

3!− z5

5!+ . . .

¶+

µz2

3!− z5

5!+ . . .

¶2+ . . .

!

=1

z3

µ1− z2

3!+ . . .

¶from which it follows that

resz=01

z2 sinh z= −1

6

and this gives the answer.

Example 24.7. Compute the integral (a > 0)Z ∞−∞

x sin ax

(1 + x2)2dx = Im

Z ∞−∞

xeiax

(1 + x2)2dx

= Im

"2πiresz=i

zeiaz

(1 + z2)2

#=

π

2ae−a.

To compute this residue, let z = i+ h and

f (z) :=zeiaz

(1 + z2)2=

zeiaz

(z − i)2 (z + i)2

then

f (i+ h) =(i+ h) eia(i+h)

h2 (h+ 2i)2=

e−a

h2(i+ h) eiah

−4 (1− ih/2)2

=e−a

−4h2(i+ h) eiah

1− ih+O (h2)

=e−a

−4h2 (i+ h)¡1 + iah+O

¡h2¢¢ ¡

1 + ih+O¡h2¢¢

=e−a

−4h2 (· · ·+ (1− a− 1)h+ . . . )

and so

resz=izeiaz

(1 + z2)2=

ae−a

4.

Example 24.8 (Summing (−1)n+1 /n2). (Sketched only very briefly!!) S69, p.245-246: #5. Let CN be the counter clockwise oriented boundary of the square

QN :=

½z ∈ C : |Re z| ≤

µN +

1

2

¶π and |Im z| ≤

µN +

1

2

¶π

¾as in Figure 24.8. Then

76 BRUCE K. DRIVER

(24.1) limN→∞

Z∂QN

1

z2 sin zdz = 0

and

(24.2)Z∂QN

1

z2 sin zdz = 2πi

"1

6+2

π2

NXn=1

(−1)nn2

#from which it follows that

(24.3)∞Xn=1

(−1)n+1n2

=π2

12.

To prove Eq. (24.1), on the part of the contours, x = ± ¡N + 12

¢π we have¯

sin

µ±µN +

1

2

¶π + iy

¶¯=

¯sin

µ±µN +

1

2

¶π

¶cos (iy) + cos

µ±µN +

1

2

¶π

¶sin (iy)

¯= |± cosh y| ≥ 1

and on the part of the contours,

y = ±µN +

1

2

¶π

we have¯sin

µx± i

µN +

1

2

¶π

¶¯=

¯sin (x) cos

µ±iµN +

1

2

¶π

¶+ cos (x) sin

µ±iµN +

1

2

¶π

¶¯=

¯sin (x) cosh

µµN +

1

2

¶π

¶± i cos (x) sinh

µµN +

1

2

¶π

¶¯∼= e(N+

12 )π ≥ 1.

Hence we have on the contour that¯1

z2 sin z

¯≤ 1¡

N + 12

¢2π2


and therefore,¯Z∂QN

1

z2 sin zdz

¯≤ 1¡

N + 12

¢2π24 (2N + 1)→ 0 as N →∞.

We now must compute the residues,

1

z2 sin z=

1

z2¡z − z3

3! +z5

5! + . . .¢ = 1

z31

1− z2

3! +z5

5! + . . .

=1

z3

Ã1 +

µz2

3!− z5

5!+ . . .

¶+

µz2

3!− z5

5!+ . . .

¶2+ . . .

!

=1

z3

µ1 +

z2

3!+ . . .

¶so that

res01

z2 sin z=1

6

while

res±nπ1

z2 sin z=

1

n2π2 cos (±nπ) = (−1)n 1

n2π2

and therefore,

Z∂QN

1

z2 sin zdz = 2πi

"1

6+

±NXn=±1

(−1)n 1

n2π2

#

= 2πi

"1

6+2

π2

NXn=1

(−1)nn2

#.

Example 24.9 (Bessel function relationship). (Skipped) #4 on p.230 (see also#10 on p. 199 )

I|z|=1

e(z+1z )dz =

I|z|=1

∞Xn=0

e1zzn

n!dz =

∞Xn=0

I|z|=1

e1zzn

n!dz

= 2πi∞Xn=0

1

n!res0

he1z zn

i= 2πi

∞Xn=0

1

n!

1

(n+ 1)!.

Recall that Bessel functions may be defined by

ez2 (w− 1

w ) =∞X

n=−∞Jn (z)w

n

78 BRUCE K. DRIVER

and so

Jm (z) =1

2πi

Z|w|=1

Ã ∞Xn=−∞

Jn (z)wn

!w−(m+1)dw

=1

2πi

Z|w|=1

ez2 (w− 1

w )w−(m+1)dw

=1

2πi

Z|w|=1

ez2w

∞Xn=0

1

n!

³z2

ńwnw−(m+1)dw

=∞Xn=0

1

n!

³z2

ń 1

2πi

Z|w|=1

ez2wwn−m−1dw

=∞Xn=0

1

n!

³z2

ńresw=0

£e

z2wwn−m−1¤

=∞Xn=0

1

n!

³z2

ń 1

m!

³z2

´m=∞Xn=0

1

n!m!

³z2

´m+n.

Writing out the contour integral explicitly we also

Jm (z) =1

2π

Z π

−πez2 (e

iθ−e−iθ)e−i(m+1)θeiθdθ

=1

2π

Z π

−πeiz sin θe−imθdθ

=1

2π

Z π

−πe−i[mθ−z sin θ]dθ

=1

2π

Z π

−πcos (mθ − z sin θ) dθ.

25. (12/3/2003) Lecture 25: More Contour Integrals

Example 25.1 (ex.abc). To evaluate sums of the form

∞Xn=1

p(n)

q (n)and

∞Xn=1

(−1)n p(n)

qQ (n)

where p and q are two polynomials with deg q ≥ deg p+ 2, one should consider theintegrals

limN→∞

Z∂QN

p (z)

q (z)cotπz dz and lim

N→∞

Z∂QN

p (z)

q (z)cscπz dz

where ∂QN is as in Example 24.8. See Problem 18 on p. 163 of Berenstein andGay, “Complex Variables: An introduction.”

Example 25.2. We wish to show

I =

Z 2π

0

cos θ

5 + 4 cos θdθ = −π

3.


(Maple gives π2 which seems to be the wrong answer here.) To this end, let z = eiθ

so that

cos θ =z + z−1

2and sin θ =

z − z−1

2iand

dz = ieiθdθ = izdθ, so that dθ =dz

iz.

We may write the integral as

I =

I|z|=1

z+z−12

5 + 4 z+z−12

dz

iz=1

2i

I|z|=1

1

z

¡z2 + 1

¢5z + 2z2 + 2

dz.

The integrand has a singular points at z = 0 and

z =−5± (4− 16)1/2

4=−5± (25− 16)1/2

4=−5± (9)1/2

4

=−5± 34

=

½−2,−1

2

¾.

Hence the answer is

I = 2πi1

2i

³res0 + res− 1

2

´"1z

¡z2 + 1

¢5z + 2z2 + 2

#= π

"1

2+

1 +¡−12¢2¡−12¢ ¡4 ¡−12¢+ 5¢

#

= π

·1

2+5/4

−32

¸= −π

3

Example 25.3. Let 0 < a < 1, we wish to compute

I =

Z ∞0

x−a

x+ 1dx =

π

sin aπ.

In order to do this we are going to consider the contour integralZCR

z−a

z + 1dz

where CR is as in Figure 11 below and z−a := e−al(z) where

l (z) = ln |z|+ iθ where z = |z| eiθ with 0 < θ < 2π.

Hence we are putting the branch cut along the real axis. By the residue calculus,ZCR

z−a

z + 1dz = 2πires−1

·z−a

z + 1

¸= 2πi (−1)−a = 2πi ¡eiπ¢−a = 2πie−iπa.

On the other hand we have, as usual,¯z−a

¯= e−aRe l(z) = e−a ln|z| = |z|−a

and hence for |z| = R we have ¯z−a

z + 1

¯∼= R−a

1 +R

so that ¯¯ZCR∩|z|=R

z−a

z + 1dz

¯¯ ≤ 2πR R−a

1 +R→ 0 as R→∞.

80 BRUCE K. DRIVER

Figure 11. A key hole contour.

Therefore we have

2πie−iπa =ZCR\|z|=R

z−a

z + 1dz =

Z R

0

x−a

x+ 1dx−

Z R

0

x−a

x+ 1dx.

Now for z = x − iε just below [0,∞) we have z−a ∼= ¡xe2πi¢−a = x−ae−2πia andhence

2πie−iπa = limR→∞

"Z R

0

x−a

x+ 1dx− e−2πia

Z R

0

x−a

x+ 1dx

#= I

¡1− e−2πia

¢.

That is

I =2πie−iπa

1− e−2πia= π

2i

eiπa − e−iπa=

π

sin aπ.

Lemma 25.4 (Jordan’s Lemma).Z π

0

e−R sin θdθ <π

R.

Proof. By symmetry and since sin θ ≤ 2π θ for θ ∈ [0, π/2], see Figure 12, we

haveZ π

0

e−R sin θdθ = 2Z π/2

0

e−R sin θdθ < 2Z π/2

0

e−R2π θdθ = − π

Re−R

2π θ|π/20 ≤ π

R.

Exercise 25.5. Show

(25.1)Z 1

−1

sinMx

xdx =

Z M

−M

sinx

xdx→ π as M →∞


32.521.510.50

1.5

1.25

1

0.75

0.5

0.25

0

x

y

x

y

Figure 12. Bounding sin θ by a straight line.

using the following method.2

(1) Show that

g(z) =

½z−1 sin z for z 6= 0

1 if z = 0

defines a holomorphic function on C.(2) Let ΓM denote the straight line path from −M to −1 along the real axis

followed by the contour eiθ for θ going from π to 2π and then followed bythe straight line path from 1 to M. Explain whyZ M

−M

sinx

xdx =

ZΓM

sin z

zdz

µ=1

2i

ZΓM

eiz

zdz − 1

2i

ZΓM

e−iz

zdz

¶.

Figure 13. The contours used in Exercise 25.5.

2In previous notes we evaluated this limit by real variable techniques based on the identitythat 1

x=R∞0 e−λxdλ for x > 0.

82 BRUCE K. DRIVER

(3) Let C+M denote the pathMeiθ with θ going from 0 to π and C−M denote thepath Meiθ with θ going from π to 2π. By deforming paths and using theCauchy integral formula, showZ

ΓM+C+M

eiz

zdz = 2πi and

ZΓM−C−M

e−iz

zdz = 0.

(4) Show (by writing out the integrals explicitly) that

limM→∞

ZC+M

eiz

zdz = 0 = lim

M→∞

ZC−M

e−iz

zdz.


26. (12/5/2003) Lecture 26: Course Review

The following two theorems summarize the main theoretical content of Math120A.

Theorem 26.1 (Analytic Functions). Let Ω ⊂o C be an open set and f ∈ C(Ω,C),then the following statements are equivalent:

(1) f ∈ H(Ω), i.e. f is analytic in Ω.(2) f(x+ iy) = u (x, y)+ iv (x, y) with u and v being continuously differentiable

functions satisfying the Cauchy Riemann equations,

fy (x+ iy) = ifx (x+ iy)

or equivalentlyuy = −vx and ux = vy.

(3)R∂T

f(z)dz = 0 for all solid triangles T ⊂ Ω.(4)

RCf (z) dz = 0 for any closed contour in Ω which is homotopic to a constant

loop.(5)

Rαf (z) dz =

Rβf (z) dz for any two contours in Ω which are homotopic in

Ω keeping the endpoints fixed.(6) For all disks D = D(z0, ρ) such that D ⊂ Ω,

(26.1) f(z) =1

2πi

I∂D

f(w)

w − zdw for all z ∈ D.

(7) For all disks D = D(z0, ρ) such that D ⊂ Ω, f(z) may be represented as aconvergent power series

(26.2) f(z) =∞Xn=0

an(z − z0)n for all z ∈ D.

In particular f ∈ C∞(Ω,C).Moreover if D is as above, we have

(26.3) f (n)(z) =n!

2πi

I∂D

f(w)

(w − z)n dw for all z ∈ D

and the coefficients an in Eq. (26.2) are given by

(26.4) an =f (n)(z0)

n!=

1

2πi

I∂D

f(w)

(w − z0)n+1dw.

We also have if A (z0, r, R) ⊂ Ω whereA (z0, r, R) := z ∈ C : r < |z − z0| < R ,

then following Laurent series converges absolutely,

f (z) =∞X

n=−∞an (z − z0)

n for all z ∈ A (z0, r, R)

where

an =1

2πi

ICρ

f (z)

(z − z0)n+1 dz.

84 BRUCE K. DRIVER

Theorem 26.2 (Residue Theorem). Suppose that f : Ω \ z1, . . . , zn → C is ananalytic function and C is a simple counter clockwise closed contour in Ω such thatC “surrounds” z1, . . . , zn , thenZ

C

f (z) dz = 2πinXi=1

reszif.

where

resz0f :=1

2πi

I|z−z0|=ε

f (z) dz = a−1

where a−1 is the coefficient of (z − z0)−1 in the Laurent series expansion of f near

z0. The following formula for computing residues is often useful:

resz=z0h (z)

g (z):=

h (z0)

g0 (z0)

provided that h and g are analytic near z0, g (z0) = 0 while g0 (z0) 6= 0.26.1. Study Guide for Math 120A Final (What you should know).

(1) C := z = x+ iy : x, y ∈ R with i2 = −1 and z = x − iy. The complexnumbers behave much like the real numbers. In particular the quadraticformula holds.

(2) |z| =px2 + y2 =

√zz, |zw| = |z| |w| , |z + w| ≤ |z| + |w| , Re z = z+z

2 ,

Im z = z−z2i , |Re z| ≤ |z| and |Im z| ≤ |z| . We also have zw = zw and

z + w = z + w and z−1 = z|z|2 .

(3) z : |z − z0| = ρ is a circle of radius ρ centered at z0.z : |z − z0| < ρ is the open disk of radius ρ centered at z0.z : |z − z0| ≥ ρ is every thing outside of the open disk of radius ρ cen-

tered at z0.(4) ez = ex (cos y + i sin y) , every z = |z| eiθ.(5) arg (z) =

©θ ∈ R : z = |z| eiθª and Arg (z) = θ if −π < θ ≤ π and z =

|z| eiθ. Notice that z = |z| ei arg(z)(6) z1/n = n

p|z|ei arg(z)n .(7) limz→z0 f (z) = L. Usual limit rules hold from real variables.(8) Mapping properties of simple complex functions(9) The definition of complex differentiable f (z) . Examples, p (z) , ez, ep(z),

1/z, 1/p (z) etc.(10) Key points of ez are is d

dz ez = ez and ezew = ez+w.

(11) All of the usual derivative formulas hold, in particular product, sum, andchain rules:

d

dzf (g (z)) = f 0 (g (z)) g0 (z)

andd

dtf (z (t)) = f 0 (z (t)) z (t) .

(12) Re z, Im z, z, are nice functions from the real - variables point of view butare not complex differentiable.

(13) Integration: Z b

a

z (t) dt :=

Z b

a

x (t) dt+ i

Z b

a

y (t) dt.


All of the usual integration rules hold, like the fundamental theorem ofcalculus, linearity and integration by parts.

(14) Be able to use the Cauchy Riemann equations to check that a function isanalytic and find harmonic conjugates

(15) You should understand and be able to use the following analytic functions:(a) ez = ex (cos y + i sin y) =

P∞n=0

1n!z

n.(b) log z = ln |z|+ i arg z and its branches:

Log (1− z) = −∞Xn=0

1

n+ 1zn+1 if |z| < 1.

(c) za and its branches: if (1 + z)α = eαLog(1+z) then

(1 + z)α =∞Xn=0

α (α− 1) . . . (α− n+ 1)

n!zn

in particular if α = −1, then1

1− z=∞Xn=0

zn.

(d) sin (z) := eiz−e−iz2i

P∞n=0

(−1)n(2n+1)!z

2n+1

(e) cos (z) := eiz+e−iz2 =

P∞n=0

(−1)n(2n)! z

2n

(f) sinh (z) := ez−e−z2 =

P∞n=0

1(2n+1)!z

2n+1

(g) cosh (z) := ez+e−z2 =

P∞n=0

1(2n)!z

2n

(h) tan (z) = sin(z)cos(z) = −i e

iz−e−izeiz+e−iz

(i) tanh (z) = sinh(z)cosh(z) =

ez−e−zez+e−z

(16) Be able to compute contour integrals by parametrizing the contour to getZC

f (z) dz =

Z b

a

f (z (t)) z (t) dt.

(17) Be able to estimate contour integrals using¯ZC

f (z) dz

¯≤ max

z∈C|f (z)| · length (C) .

(18) Be able to compute contour integrals using the fundamental theorem ofcalculus: if f is analytic on a neighborhood of a contour C, thenZ

C

f 0 (z) dz = f(Cend)− f (Cbegin) .

(19) Be able to compute simple Taylor series and Laurent series expansions ofa function f centered at a point z0 ∈ C. Hint: If z0 6= 0, write z = z0 + hand then do the expansion in h about h = 0. At the end replace h by z−z0.

(20) Be able to compute residues and use the residue theorem for computingcontour integrals.

(21) Be able to use complex techniques to compute real integrals as have ap-peared on the homework problems.

86 BRUCE K. DRIVER

27. For Those Interested: Theory Skipped in Lectures

27.1. Differentiating and integrating a sum of analytic functions. We nowrestate and prove the differentiating and integrating a sum of analytic functionsTheorem 20.1.

Theorem 27.1 (Differentiating and integrating a sum of analytic functions). Sup-pose that fn : Ω→ C is a sequence of analytic functions such that

|fn (z)| ≤Mn for all n ∈ N and z ∈ Cwhere

P∞n=1Mn <∞. Then

(1) If C is any contour in Ω, we haveZC

F (z) dz =∞Xn=1

ZC

fn (z) dz.

(2) The function F (z) :=P∞

n=1 fn (z) is an analytic.(3) F 0 (z) =

P∞n=1 f

0n (z) and in fact

(27.1) F (k) (z) =∞Xn=1

f (k)n (z) for all k ∈ N0 and z ∈ Ω.

Part of the assertion here is that all sums appearing are absolutely convergent.

Proof.

(1) Since∞Xn=1

¯ZC

fn (z) dz

¯≤∞Xn=1

Mn (C) <∞

where (C) is the length of C, the sumP∞

n=1

RCfn (z) dz is absolutely

convergent. Moreover¯¯ZC

F (z) dz −NXn=1

ZC

fn (z) dz

¯¯ =

¯¯ZC

"F (z)−

NXn=1

fn (z)

#dz

¯¯ ≤ εN (C)

where

εN :=maxC

¯¯F (z)−

NXn=1

fn (z)

¯¯ ≤ maxC

¯¯∞X

n=N+1

fn (z)

¯¯

≤∞X

n=N+1

Mn → 0 as N →∞.

(2) Suppose that T is a solid triangle inside of Ω, then by item 1.,Z∂T

F (z) dz =∞Xn=1

Z∂T

fn (z) dz = 0

where the last equality is a consequence of the Cauchy Goursat Theoremor the converse to Morera’s theorem. It now follow by an application ofMorera’s theorem that F is analytic on Ω. (Item 2. will also be proved inthe course of the proof of item 3.)


(3) Since complex differentiability is a local assertion, let us fix z0 ∈ Ω andρ > 0 such that D (z0, ρ) ⊂ Ω and let r = dist(z0, ∂Ω) > ρ. Then by theCauchy estimate in Corollary 18.7 with n = 1, we learn

|f 0n (z)| ≤1

r − ρMn for all z ∈ D (z0, ρ) .

We now suppose z ∈ D (z0, ρ) and h ∈ C\ 0 with |h| < ρ− |z| . Using thedefinition of the derivative and properties of the sum,

F (z + h)− F (z)

h−∞Xn=1

f 0n (z) =∞Xn=1

·fn (z + h)− fn (z)

h− f 0n (z)

¸.

By the fundamental theorem of calculus and the chain rule,

fn (z + h)− fn (z) =

Z 1

0

d

dtfn (z + th) dt = h

Z 1

0

f 0n (z + th) dt

which implies¯fn (z + h)− fn (z)

h− f 0n (z)

¯=

¯Z 1

0

[f 0n (z + th)− f 0n (z)] dt¯

≤Z 1

0

|f 0n (z + th)− f 0n (z)| dt ≤2

r − ρMn.

Therefore for any N ∈ N we have¯¯F (z + h)− F (z)

h−∞Xn=1

f 0n (z)

¯¯ ≤

∞Xn=1

¯fn (z + h)− fn (z)

h− f 0n (z)

¯

≤NXn=1

¯fn (z + h)− fn (z)

h− f 0n (z)

¯+2

ρ

∞Xn=N+1

Mn.

So letting h→ 0 in this expression shows

limh→0

¯¯F (z + h)− F (z)

h−∞Xn=1

f 0n (z)

¯¯ ≤ 2ρ

∞Xn=N+1

Mn → 0 and N →∞.

This procedure may be repeated to prove Eq. (27.1). Since z0 ∈ Ω wasarbitrary, the proof is complete.

27.2. The Basic Theory of Power Series.

Lemma 27.2 (Root and Ratio Test). Suppose zn∞n=0 ⊂ C and

ρ := limn→∞

np|zn| or ρ := lim

n→∞

¯zn+1zn

¯exists

then

(1) If ρ < 1 thenP∞

n=0 |zn| <∞ and henceP∞

n=0 zn is convergent.(2) If ρ > 1, limn→∞ |zn| =∞ and

P∞n=0 zn is divergent.

(3) If ρ = 1, the test fails and you have to work harder.

Proof.(1) Suppose ρ < 1 and let ρ < r < 1.

88 BRUCE K. DRIVER

(a) Suppose first that ρ := limn→∞ np|zn|, then n

p|zn| ≤ r for large nand hence

|zn| ≤ rn for large n, say n ≥ N.

SinceP∞

N |zn| ≤P∞

N rn ≤ 11−r < ∞, the sum is absolutely conver-

gent.(b) Suppose now that ρ := limn→∞

¯zn+1zn

¯. Then again for n ≥ N for some

N we have¯zn+1zn

¯≤ r i.e. |zn+1| ≤ r |zn| for all n ≥ N.

This then implies

|zN+n| ≤ r |zN+n−1| ≤ r2 |zN+n−2| ≤ · · · ≤ rn |zN | .So again

∞XN

|zn| ≤ |zN |∞XN

rn ≤ |zN | 1

1− r<∞

and the original sum is absolutely convergent.(2) Suppose ρ > 1 and let ρ > r > 1.

(a) Suppose first that ρ := limn→∞ np|zn|, then n

p|zn| ≥ r for large nand hence

|zn| ≥ rn for large n

and hence limn→∞ |zn| =∞ and the series must diverge.

(b) Suppose now that ρ := limn→∞¯zn+1zn

¯. Then again for n ≥ N for some

N we have¯zn+1zn

¯≥ r i.e. |zn+1| ≥ r |zn| for all n ≥ N.

This then implies

|zN+n| ≥ r |zN+n−1| ≥ r2 |zN+n−2| ≥ · · · ≥ rn |zN | .So again limn→∞ |zn| =∞ and the series diverge.

Definition 27.3. Given z0 ∈ C and an∞n=0 ⊂ C, the series of the form∞Xn=0

an (z − z0)n

is called a power series. If z0 = 0 we call it a Maclaurin series, i.e. a series ofthe form ∞X

n=0

anzn.

To each power series,P∞

n=0 an (z − z0)n , let

r := sup

(|z − z0| : z ∈ C and

∞Xn=0

an (z − z0)n exists

).

The number r ≥ 0 is called the radius of convergence of the series.


Proposition 27.4. If r is the radius of convergence of a power series

(27.2)∞Xn=0

an (z − z0)n

then:

(1) If |z − z0| < r, the series converges.(2) If |z − z0| > r, the series diverges.(3) If

µ = limn→∞

np|an| or µ = lim

n→∞

¯an+1an

¯exist

then r = 1µ .

Proof. For simplicity of exposition we will assume that z0 = 0.(1) If w ∈ C\ 0 is a point such that P∞n=0 anwn exists then, with λ = |w| ,

limn→∞ |anw

n| = lim |an|λn = 0.In particular for large n we have |an|λn ≤ 1 or |an| ≤ λ−n. Hence if |z| < λ,then

|anzn| ≤µ |z|

λ

¶nfor large n and hence

P∞n=0 |anzn| < ∞ by comparison with a geometric

series. This proves item 1.(2) If |w| > r and

P∞n=0 anw

n were to exists, this would violate the definitionof r.

(3) If we apply the root test or the ratio test to the series in Eq. (27.2) wewould learn

ρ := limn→∞

n

q|an (z − z0)

n| = µ |z − z0| or

ρ := limn→∞

¯an+1 (z − z0)

n+1¯

|an (z − z0)n| = µ |z − z0|

and in either case we would learn that the series in Eq. (27.2) converges ifρ < 1 and diverges if ρ > 1 and these later conditions are equivalent to

|z − z0| < 1

µand |z − z0| > 1

µ.

It follows from this that r = 1µ in this case.

Using these results and our differentiation Theorem 20.1 we get the followingcorollary.

Theorem 27.5 (Power Series Integration and Differentiation). Suppose that

S (z) :=∞Xn=0

an (z − z0)n for z ∈ D := D (z0, r)

where r is the radius of convergence of the series which is assumed to be positive.Then:

90 BRUCE K. DRIVER

(1) S is analytic on D and

(27.3) an =1

n!S(n) (z0) for all n ∈ N0.

(2) The derivative S is given by

(27.4) S0 (z) =∞Xn=1

nan (z − z0)n−1 .

and more generally,

S(k) (z) =∞Xn=0

n(n− 1) . . . (n− k + 1)an (z − z0)n−k for all z ∈ D (z0, r) .

(3) If C is a contour in D then(27.5)Z

C

S (z) dz =∞Xn=0

an

ZC

(z − z0)ndz =

∞Xn=0

ann+ 1

h(zf − z0)

n+1 − (zi − z0)n+1

iwhere zi and zf are the initial and final points of C respectively. In partic-ular if zi = z0, thenZ

[z0,w]

S (z) dz =∞Xn=0

ann+ 1

(w − z0)n+1

.

Proof. This all follows from Theorem 20.1 and our discussions about powerseries in Proposition 27.4 and its proof.

Corollary 27.6 (Removable singularities). Let Ω ⊂o C, z0 ∈ Ω and f ∈ H(Ω \z0). If lim supz→z0 |f(z)| < ∞, i.e. sup

0<|z−z0|<|f(z)| < ∞ for some > 0, then

limz→z0

f(z) exists. Moreover if we extend f to Ω by setting f(z0) = limz→z0

f(z), then

f ∈ H(Ω).

Proof. Set

g(z) =

½(z − z0)

2f(z) for z ∈ Ω \ z00 for z = z0

.

Then g0(z0) exists and is equal to zero. Therefore g0(z) exists for all z ∈ Ω andhence g ∈ H(Ω).We may now expand g into a power series using g(z0) = g0(z0) = 0

to learn g(z) =∞Pn=2

an(z − z0)n which implies

f(z) =g(z)

(z − z0)2=∞Xn=0

an(z − z0)n−2 for 0 < |z − z0| <

Therefore, limz→z0 f(z) = a2 exists. Defining f(z0) = a2 we have f(z) =∞Pn=0

an(z−z0)

n−2 for z near z0. This shows that f is holomorphic in a neighborhood of z0 andsince f was already holomorphic away from z0, f ∈ H(Ω).

Definition 27.7. We say that Ω is a region if Ω is a connected open subset of C.


Corollary 27.8 (Analytic Continuation). Let Ω be a region and f ∈ H(Ω)and Z(f) = f−1(0) denote the zero set of f. Then either f ≡ 0 or Z(f)has no accumulation points in Ω. More generally if f, g ∈ H(Ω) and the setz ∈ Ω : f(z) = g(z) has an accumulation point in Ω, then f ≡ g.

Proof. The second statement follows from the first by considering the functionf −g. For the proof of the first assertion we will work strictly in Ω with the relativetopology.Let A denote the set of accumulation points of Z(f) (in Ω). By continuity of

f, A ⊂ Z(f) and A is a closed3 subset of Ω with the relative topology. The proofis finished by showing that A is open and thus A = ∅ or A = Ω because Ω isconnected.Suppose that z0 ∈ A, and express f(z) as its power series expansion

f(z) =∞Xn=0

an(z − z0)n

for z near z0. Since 0 = f(z0) it follows that a0 = 0. Let zk ∈ Z(f) \ z0 such thatlim zk = z0. Then

0 =f(zk)

zk − z0=∞Xn=1

an(zk − z0)n−1 → a1 as k →∞

so that f(z) =P∞

n=2 an(z − z0)n. Similarly

0 =f(zk)

(zk − z0)2 =

∞Xn=2

an(zk − z0)n−2 → a2 as k →∞

and continuing by induction, it follows that an ≡ 0, i.e. f is zero in a neighborhoodof z0.

Definition 27.9. For z ∈ C, let

cos(z) =eiz + eiz

2and sin(z) =

eiz − eiz

2i.

Exercise 27.10. Show the these formula are consistent with the usual definitionof cos and sin when z is real. Also shows that the addition formula in Exercise31.15 are valid for θ, α ∈ C. This can be done with no additional computations bymaking use of Corollary 27.8.

Exercise 27.11. Let

f(z) :=1√2π

ZRexp(−1

2x2 + zx)dm(x) for z ∈ C.

Show f(z) = exp( 12z2) using the following outline:

(1) Show f ∈ H(Ω).(2) Show f(z) = exp( 12z

2) for z ∈ R by completing the squares and using thetranslation invariance of m. Also recall that you have proved in the firstquarter that f(0) = 1.

(3) Conclude f(z) = exp( 12z2) for all z ∈ C using Corollary 27.8.

3Recall that x ∈ A iff V 0x ∩ Z 6= ∅ for all x ∈ Vx ⊂o C where V 0

x := Vx \ x . Hence x /∈ A iffthere exists x ∈ Vx ⊂o C such that V 0

x ∩Z = ∅. Since V 0x is open, it follows that V 0x ⊂ Ac and thus

Vx ⊂ Ac. So Ac is open, i.e. A is closed.

92 BRUCE K. DRIVER

27.3. Partial Fractions. Consider writing q(z)p(z) in partial fraction form. Here we

assume deg q < deg p, for otherwise we would divide to make it so. Now factp (z) =

Qni=1 (z − zi)

ki we wish to write

q (z)Qni=1 (z − zi)

ki=

nXi=1

kiXj=1

aij1

(z − zi)ki−j+1 .

Multiplying this equation through by p (z) shows we must solve

q (z) =nXi=1

kiXj=1

aij (z − zi)j−1

nYl 6=i(z − zl)

kl .

Noting that k := deg p =P

ki, the question comes down to whether the functions

β :=

pij (z) := (z − zi)j−1

nYl6=i(z − zl)

kl : i = 1, . . . , n and j = 1, . . . , ki

form a basis for the polynomials of degree k − 1. This space has dimension k andthere are k elements in β. So to finish the proof, we need only show that β is alinearly independent set. Suppose that

(27.6) F (z) :=nXi=1

kiXj=1

aijpij (z) = 0.

Evaluating this expression at z1 shows

0 =nXi=1

kiXj=1

aijpij (z1) = a11

nYl6=1(z1 − zl)

kl

which implies a11 = 0. Similarly by evaluating at zi we learn that ai1 = 0 for all iand we are done if ki = 1 for all i. So we are left to consider

0 =nXi=1

Xj≥2

aijpij (z) .

This expression will have a common factor ofQ

i:ki>1(z − zi) which when factored

out, leaves us to consider

0 =nXi=1

Xj≥2

aij pij (z)

where

pij (z) =pij (z)Q

i:ki>1(z − zi)

.

In this way we have reduced the maximum ki appearing by 1. Hence we maycomplete the proof by induction on max ki : i = 1, . . . , n .Example 27.12. Suppose p (z) = (z − z1) (z − z2)

2(z − z3)

3 where now K :=max ki : i = 1, 2, 3 = 3. Then we are considering

0 = A (z − z2)2 (z − z3)

3 +B (z − z1) (z − z2) (z − z3)3

+ C (z − z1) (z − z3)3 +D (z − z1) (z − z2)

2 (z − z3)2

+E (z − z1) (z − z2)2 (z − z3) + F (z − z1) (z − z2)

2 .


Evaluating at z = z1 implies, A = 0 and z = z2 that C = 0 and at z = z3 thatF = 0. So we are left to consider

0 = B (z − z1) (z − z2) (z − z3)3 +D (z − z1) (z − z2)

2 (z − z3)2

+E (z − z1) (z − z2)2 (z − z3)

from which we can factor out (z − z1) (z − z2) (z − z3) to find

0 = B (z − z3)2+D (z − z2) (z − z3) +E (z − z2)

and now K := max ki : i = 1, 2, 3 = 2. Evaluating this at z = z1 and z = z2implies that E = D = 0, so that

0 = B (z − z3)2

and we may factor out (z − z3) to get

0 = B (z − z3) .

Evaluating this at any point other than z3 implies B = 0.

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